# 20588537-Chuong7-Phuong-Trinh-Luong-Giac-Chua-Can-Thuc-Va-Gia-Tri-Tuyet-Doi

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```							CHÖÔNG VII

PHÖÔNG TRÌNH LÖÔÏ N G GIAÙ C CHÖÙ A CAÊ N VAØ PHÖÔNG TRÌNH LÖÔÏ N G GIAÙ C CHÖÙ A GIAÙ TRÒ TUYEÄT ÑOÁI
A) PHÖÔNG TRÌNH LÖÔÏ N G GIAÙ C CHÖÙ A CAÊ N Caù c h giaû i : AÙ p duï n g caù c coâ n g thöù c ⎧A ≥ 0 A = B⇔⎨ ⇔ ⎩A = B

⎧B ≥ 0 A =B⇔⎨ 2 ⎩A = B Ghi chuù : Do theo phöông trình chænh lyù ñaõ boû phaà n baá t phöông trình löôï n g giaù c neâ n ta xöû lyù ñieà u kieä n B ≥ 0 baè n g phöông phaù p thöû laï i vaø chuù n g toâ i boû caù c baø i toaù n quaù phöù c taï p .

⎧B ≥ 0 ⎨ ⎩A = B

Baø i 138 : Giaû i phöông trình

5 cos x − cos 2x + 2 sin x = 0 ( *)

( *) ⇔

5 cos x − cos 2x = −2 sin x

⎧sin x ≤ 0 ⇔⎨ 2 ⎩5 cos x − cos 2x = 4 sin x ⎧sin x ≤ 0 ⎪ ⇔⎨ 2 2 ⎪5 cos x − 2 cos x − 1 = 4 1 − cos x ⎩ ⎧sin x ≤ 0 ⇔⎨ 2 ⎩2 cos x + 5 cos x − 3 = 0 ⎧sin x ≤ 0 ⎪ ⇔⎨ 1 ⎪cos x = 2 ∨ cos x = −3 ( loaïi ) ⎩ ⎧sin x ≤ 0 ⎪ ⇔⎨ π ⎪ x = ± 3 + k2π, k ∈ ⎩ π ⇔ x = − + k2π, k ∈ 3

(

)

(

)

Baø i 139 : Giaû i phöông trình sin3 x + cos3 x + sin3 x cot gx + cos3 xtgx = 2 sin 2x

Ñieà u kieä n : ⎧cos x ≠ 0 ⎪ ⎨sin x ≠ 0 ⇔ ⎪sin 2x ≥ 0 ⎩

⎧sin 2x ≠ 0 ⇔ sin 2x > 0 ⎨ ⎩sin 2x ≥ 0

Luù c ñoù : ( *) ⇔ sin3 x + cos3 x + sin2 x cos x + cos2 x sin x = 2 sin 2x

⇔ sin2 x ( sin x + cos x ) + cos2 x ( cos x + sin x ) = 2sin 2x
⇔ ( sin x + cos x ) sin 2 x + cos2 x = 2 sin 2x

(

)

⎧sin x + cos x ≥ 0 ⎪ ⇔⎨ 2 ⎪( sin x + cos x ) = 2 sin 2x ⎩
⎧ π⎞ ⎛ ⎧ π⎞ ⎛ ⎪ 2 sin ⎜ x + ⎟ ≥ 0 ⎪sin ⎜ x + ⎟ ≥ 0 4⎠ ⇔⎨ ⇔⎨ 4⎠ ⎝ ⎝ ⎪1 + sin 2x = 2 sin 2x ⎪sin 2x = 1 ( nhaän do sin 2x > 0 ) ⎩ ⎩ ⎧ π⎞ ⎧ π⎞ ⎛ ⎛ ⎪sin ⎜ x + 4 ⎟ ≥ 0 ⎪sin ⎜ x + 4 ⎟ ≥ 0 ⎪ ⎪ ⎝ ⎠ ⎝ ⎠ ⇔⎨ ⇔⎨ ⎪ x = π + kπ, k ∈ ⎪ x = π + m2π ∨ x = 5π + m2π ( loaïi ) , m ∈ ⎪ ⎪ ⎩ 4 ⎩ 4 4 π ⇔ x = + m2π, m ∈ 4

Baø i 140 : Giaû i phöông trình

π⎞ ⎛ 1 + 8 sin 2x. cos2 2x = 2 sin ⎜ 3x + ⎟ ( *) 4⎠ ⎝

⎧ π⎞ ⎛ ⎪sin ⎜ 3x + 4 ⎟ ≥ 0 ⎪ ⎝ ⎠ Ta coù : (*) ⇔ ⎨ ⎪1 + 8 sin 2x cos2 2x = 4 sin2 ⎛ 3x + π ⎞ ⎜ ⎟ ⎪ 4⎠ ⎝ ⎩ ⎧ π⎞ ⎛ ⎪sin ⎜ 3x + 4 ⎟ ≥ 0 ⎪ ⎝ ⎠ ⇔⎨ ⎪1 + 4 sin 2x (1 + cos 4x ) = 2 ⎡1 − cos( 6x + π ) ⎤ ⎢ ⎪ 2 ⎥ ⎣ ⎦ ⎩ ⎧ π⎞ ⎛ ⎪sin ⎜ 3x + ⎟ ≥ 0 4⎠ ⇔⎨ ⎝ ⎪1 + 4 sin 2x + 2 ( sin 6x − sin 2x ) = 2 (1 + sin 6x ) ⎩
⎧ π⎞ ⎧ π⎞ ⎛ ⎛ ⎪sin ⎜ 3x + 4 ⎟ ≥ 0 ⎪sin ⎜ 3x + 4 ⎟ ≥ 0 ⎪ ⎪ ⎝ ⎠ ⎝ ⎠ ⇔⎨ ⇔⎨ ⎪sin 2x = 1 ⎪ x = π + kπ ∨ x = 5π + kπ, k ∈ ⎪ ⎪ ⎩ 2 ⎩ 12 12

π⎞ ⎛ So laï i vôù i ñieà u kieä n sin ⎜ 3x + ⎟ ≥ 0 4⎠ ⎝ π •Khi x = + kπ thì 12 π⎞ ⎛ ⎛π ⎞ sin ⎜ 3x + ⎟ = sin ⎜ + 3kπ ⎟ = cos kπ 4⎠ ⎝ ⎝2 ⎠ ⎡1 , ( neáu k chaün ) ( nhaän ) =⎢ ⎢ −1 , ( neáu k leû ) ( loaïi ) ⎣ 5π • Khi x = + kπ thì 12 π⎞ ⎛ ⎛ 3π ⎞ ⎛ π ⎞ sin ⎜ 3x + ⎟ = sin ⎜ + 3kπ ⎟ = sin ⎜ − + kπ ⎟ 4⎠ ⎝ ⎝ 2 ⎠ ⎝ 2 ⎠

⎡ −1 , neáu k chaün ( loaïi ) =⎢ ⎢1 , neáu k leû ( nhaän ) ⎣ π 5π + m2π ∨ x = + ( 2m + 1) π, m ∈ Do ñoù ( *) ⇔ x = 12 12
1 − sin 2x + 1 + sin 2x = 4 cos x ( * ) sin x

Baø i 141 : Giaû i phöông trình

Luù c ñoù : ( *) ⇔ 1 − sin 2x + 1 + sin 2x = 2 sin 2x ( hieå n nhieâ n sinx = 0 khoâ n g laø nghieä m , vì sinx =0 thì VT = 2, VP = 0 ) ⎧ ⎪2 + 2 1 − sin2 2x = 4 sin2 2x ⇔⎨ ⎪sin 2x ≥ 0 ⎩ ⎧ 1 − sin2 2x = 2 sin2 2x − 1 ⎪ ⇔⎨ ⎪sin 2x ≥ 0 ⎩

⇔

⇔

⇔

⇔

⎧1 − sin 2 2x = 4 sin4 2x − 4 sin2 2x + 1 ⎪ 1 ⎪ 2 ⎨sin 2x ≥ 2 ⎪ ⎪sin 2x ≥ 0 ⎩ ⎧sin 2 2x 4 sin 2 2x − 3 = 0 ⎪ ⎨ 1 ⎪sin 2x ≥ 2 ⎩ ⎧ 3 − 3 ∨ sin 2x = ⎪sin 2x = ⎪ 2 2 ⎨ ⎪sin 2x ≥ 2 ⎪ 2 ⎩ 3 sin 2x = 2

(

)

π 2π + k2π ∨ 2x = + k2π, k ∈ 3 3 π π ⇔ x = + kπ ∨ x = + kπ, k ∈ 6 3 Chuù yù : Coù theå ñöa veà phöông trình chöù a giaù trò tuyeä t ñoá i ⎧sin x ≠ 0 ( *) ⇔ ⎪ ⎨ ⎪ cos x − sin x + cos x + sin x = 2 sin 2x ⎩ ⇔ cos x − sin x + cos x + sin x = 2 sin 2x ⇔ 2x =
Baø i 142 : Giaû i phöông trình sin x + 3 cos x + sin x + 3 cos x = 2 ( * )
π 3 cos x Ñaët t = sin x + 3 cos x = sin x + π cos 3 π⎞ π⎞ 1 ⎛ ⎛ ⇔t= sin ⎜ x + ⎟ = 2 sin ⎜ x + ⎟ π 3⎠ 3⎠ ⎝ ⎝ cos 3 ( *) thaønh t + t = 2 sin

⇔

t = 2−t

Do ñoù ( * )

⎧2 − t ≥ 0 ⎧t ≤ 2 ⇔⎨ ⇔⎨ 2 2 ⎩t = 4 − 4t + t ⎩t − 5t + 4 = 0 ⎧t ≤ 2 ⇔⎨ ⇔ t =1 ⎩t = 1 ∨ t = 4

π⎞ 1 π π π 5π ⎛ ⇔ sin ⎜ x + ⎟ = ⇔ x + = + k2π hay x + = + k2π, k ∈ 3⎠ 2 3 6 3 6 ⎝ π π ⇔ x = − + k2π ∨ x = + k2π, k ∈ 6 2

Baø i 143 : Giaû i phöông trình 3 tgx + 1 ( sin x + 2 cos x ) = 5 ( sin x + 3 cos x ) ( *) Chia hai veá cuû a (*) cho cos x ≠ 0 ta ñöôï c ( *) ⇔ 3 tgx + 1 ( tgx + 2) = 5 ( tgx + 3) Ñaët u =
tgx + 1 vôùi u ≥ 0

Thì u 2 − 1 = tgx

(*) thaøn h 3u u 2 + 1 = 5 u 2 + 2

(

)

(

)

⇔ 3u 3 − 5u 2 + 3u − 10 = 0 ⇔ ( u − 2 ) ( 3u 2 + u + 5 ) = 0
⇔ u = 2 ∨ 3u 2 + u + 5 = 0 ( voâ nghieäm )

Do ñoù

( *) ⇔

tgx + 1 = 2

⇔ tgx + 1 = 4
π π⎞ ⎛ ⇔ tgx = 3 = tgα ⎜ vôùi − < α < ⎟ ⇔ x = α + k π , k ∈ 2 2⎠ ⎝

Baø i 144 : Giaû i phöông trình

(

1 − cos x + cos x cos 2x =

)

( *) ⇔ (

1 − cos x + cos x cos 2x = sin 2x cos 2x

)

1 sin 4x ( *) 2

⎧cos x ≥ 0 ⇔⎨ hay ⎩cos 2x = 0
⎧cos x ≥ 0 ⎪ ⇔⎨ π ⎪2x = 2 + kπ, k ∈ ⎩

1 − cos x + cos x = sin 2x

⎧cos x ≥ 0 ⎪ hay ⎨sin 2x ≥ 0 ⎪ 2 ⎩1 + 2 ( 1 − cos x)cosx = sin 2x ⎧cos x ≥ 0 ⎧cos x ≥ 0 ⎪ ⎪ hay ⎨sin 2x ≥ 0 ⇔⎨ π π ⎪x = 4 + k 2 , k ∈ ⎪ 2 ⎩ ⎩1 + 2 ( 1 − cos x)cosx = sin 2x ( VT ≥ 1 ≥ VP ) ⎧cos x ≥ 0 cos x ≥ 0 ⎪sin 2x ≥ 0 ⎧ ⎪ ⎪ ⇔⎨ hay ⎨ 2 π 5π ⎪ x = ± 4 + hπ hay x = ± 4 + hπ, h ∈ ⎪sin 2x = 1 ⎩ ⎪ ⎩(1 − cos x ) cos x = 0 π ⇔ x = ± + hπ, h ∈ 4 ⎧sin 2x = 1 ⎧sin 2x = 1 hay ⎨ hay ⎨ ⎩cos x = 0 ( ⇒ sin 2x = 0 ) ⎩cos x = 1 ( ⇒ sin x = 0 ⇒ sin 2x = 0 ) π ⇔ x = ± + hπ, h ∈ 4

Baø i 145 : Giaû i phöông trình sin3 x (1 + cot gx ) + cos3 x (1 + tgx ) = 2 sin x cos x ( *)

( *) ⇔ sin3 x ⎛ ⎜

⇔ ( sin x + cos x ) sin 2 x + cos2 x = 2 sin x cos x

sin x + cos x ⎞ ⎛ cos x + sin x ⎞ 3 ⎟ + cos x ⎜ ⎟ = 2 sin x cos x sin x cos x ⎝ ⎠ ⎝ ⎠

(

)

⎧sin x + cos x ≥ 0 ⇔⎨ ⎩1 + sin 2x = 2 sin 2x
⎧ π⎞ ⎛ ⎪sin ⎜ x + 4 ⎟ ≥ 0 ⎧sin x + cos x ≥ 0 ⎪ ⎝ ⎠ ⇔⎨ ⇔⎨ ⎩sin 2x = 1 ⎪ x = π + kπ, k ∈ ⎪ 4 ⎩

⎧ π⎞ ⎛ ⎪sin ⎜ x + 4 ⎟ ≥ 0 ⎪ ⎝ ⎠ ⇔⎨ ⎪ x + π = π + kπ, k ∈ ⎪ 4 2 ⎩ ⎧ π⎞ ⎛ ⎪sin ⎜ x + 4 ⎟ ≥ 0 ⎪ ⎝ ⎠ ⇔⎨ ⎪ x + π = π + h2π hay x + π = 3π + h2π, h ∈ ⎪ ⎩ 4 2 4 2 π ⇔ x = + h2π, h ∈ 4

Baø i 146 : Giaû i phöông trình

cos 2x + 1 + sin 2x = 2 sin x + cos x ( *)

π⎞ ⎛ Ñieà u kieä n cos 2x ≥ 0 vaø sin ⎜ x + ⎟ ≥ 0 4⎠ ⎝

Luù c ñoù : ( *) ⇔

cos2 x − sin 2 x +
2

( cos x + sin x )

2

= 2 cos x + sin x
2

⇔ cos2 x − sin 2 x + ( cos x + sin x ) + 2 cos 2x

⇔ cos x ( cos x + sin x ) + ( sin x + cos x ) cos 2x

( cos x + sin x ) = 4 ( sin x + cos x ) = 2 ( sin x + cos x )

⎡sin x + cos x = 0 ⇔⎢ ⎣cos x + cos 2x = 2 ⎡ tgx = −1 ⇔⎢ ⎢ cos 2x = 2 − cos x ( * *) ⎣ ⎡ tgx = −1 ⇔⎢ 2 ⎣cos 2x = 4 − 4 cos x + cos x ⇔ tgx = −1 ∨ cos2 x + 4 cos x − 5 = 0 ⇔ tgx = −1 ∨ cos x = 1 ∨ cos x = −5 ( loaïi )

π + kπ ∨ x = k2π, k ∈ 4 π ⎛ π⎞ Thöû laï i : • x = − + kπ thì cos 2x = cos ⎜ − ⎟ = 0 ( nhaän ) 4 ⎝ 2⎠ π⎞ ⎛ Vaø sin ⎜ x + ⎟ = sin kπ = 0 ( nhaän ) 4⎠ ⎝ • x = k2π thì cos 2x = 1 ( nhaän ) ⇔x=−
π⎞ π ⎛ vaø cos ⎜ x + ⎟ = cos > 0 ( nhaän ) 4⎠ 4 ⎝ π Do ñoù (*) ⇔ x = − + kπ ∨ x = k2π, k ∈ 4 Chuù yù : Taï i (**) coù theå duø n g phöông trình löôï n g giaù c khoâ n g möï c

( * *) ⇔ ⎪ ⎨

⎧cos x = 1 ⇔⎨ ⇔ x = 2kπ, k ∈ ⎩sin x + cos x ≥ 0 Caù c h khaù c

⎧cos x = 1 ⎪ ⇔ ⎨cos 2x = 2 cos2 x − 1 = 1 ⎪sin x + cos x ≥ 0 ⎩

⎧cos x + cos 2x = 2 ⎪sin x + cos x ≥ 0 ⎩

( *) ⇔
⇔

cos2 x − sin 2 x +

( cos x + sin x )

2

= 2 cos x + sin x
2

(cos x + sin x).(cos x − sin x ) +

( cos x + sin x )

= 2 cos x + sin x

⎧cos x + sin x > 0 ⎪ ⇔ cos x + sin x = 0 hay ⎨ ⎪ cos x − sin x + ⎩ ⎧cos x + sin x > 0 ⎪ ⇔ tgx = − 1 hay ⎨ ⎪2 cos x + 2 cos 2x = 4 ⎩

( cos x + sin x ) = 2

⎧cos x + sin x > 0 ⎪ ⇔ tgx = − 1 hay ⎨ ⎪cos x + cos 2x = 2 ⎩ ⎧cos x = 1 π ⇔ x = − + kπ, k ∈ hay ⎨ 4 ⎩cos 2x = 1 π ⇔ x = − + kπ hay x = 2kπ, k ∈ 4 ( nhaä n xeù t : khi cosx =1 thì sinx = 0 vaø sinx + cosx = 1 > 0 )

1. Giaû i phöông trình : a/ 1 + sin x + cos x = 0 4x − cos2 x cos 3 =0 b/ 1 − tg 2 x d/

BAØI TAÄP

c/ sin x + 3 cos x = 2 + cos 2x + 3 sin 2x
sin 2 x − 2 sin x + 2 = 2 sin x − 1 3tgx e/ 2 3 sin x = − 3 2 sin x − 1

sin2 2x + cos4 2x − 1 =0 f/ sin cos x g/ 8 cos 4x cos2 2x + 1 − cos 3x + 1 = 0
h/

sin x + sin x + sin2 x + cos x = 1

k/ 5 − 3sin 2 x − 4 cos x = 1 − 2 cos x l/ cos 2x = cos2 x 1 + tgx 2. Cho phöông trình : 1 + sin x + 1 − sin x = m cos x (1) a/ Giaû i phöông trình khi m = 2 b/ Giaû i vaø bieä n luaä n theo m phöông trình (1) 3. Cho f(x) = 3cos 6 2x + sin 42x + cos4x – m a/ Giaû i phöông trình f(x) = 0 khi m = 0 b/ Cho g ( x ) = 2 cos2 2x 3 cos2 2x + 1 . Tìm taá t caû caù c giaù trò m ñeå phöông trình f(x) = g(x) coù nghieä m . ( ÑS : 1 ≤ m ≤ 0 ) 4. Tìm m ñeå phöông trình sau coù nghieä m 1 + 2 cos x + 1 + 2sin x = m

(ÑS :

1+ 3 ≤ m ≤ 2 1+ 2

)

B) PHÖÔNG TRÌNH LÖÔÏ N G GIAÙ C CHÖÙ A CAÙ C TRÒ TUYEÄ T ÑOÁ I Caù ch giaû i : 1/ Môû giaù trò tuyeä t ñoá i baè n g ñònh nghóa 2/ AÙ p duï n g • A = B ⇔ A = ±B

⎧B ≥ 0 ⎧B ≥ 0 ⎧A ≥ 0 ⎧A < 0 •A =B⇔⎨ ⇔⎨ 2 ⇔⎨ ∨⎨ 2 ⎩ A = ±B ⎩ A = B ⎩ A = −B ⎩A = B
Baø i 147 : Giaû i phöông trình cos 3x = 1 − 3 sin 3x ( *)

( *) ⇔ ⎪ ⎨

⎧1 − 3 sin 3x ≥ 0

1 ⎧ ⎪sin 3x ≤ 3 ⇔⎨ ⎪4 sin 2 3x − 2 3 sin 3x = 0 ⎩

2 2 ⎪cos 3x = 1 − 2 3 sin 3x + 3sin 3x ⎩ 1 ⎧ ⎪sin 3x ≤ 3 ⇔⎨ ⎪1 − sin 2 3x = 1 − 2 3 sin 3x + 3 sin 2 3x ⎩

1 ⎧ ⎪sin 3x ≤ 3 ⎪ ⇔⎨ ⎪sin 3x = 0 ∨ sin 3x = 3 ⎪ ⎩ 2 ⇔ sin 3x = 0

⇔x=

kπ ,k ∈ 3

Baø i 148 : Giaû i phöông trình 3sin x + 2 cos x − 2 = 0 ( * )

( *) ⇔ 2 cos x

= 2 − 3sin x

⎧2 − 3sin x ≥ 0 ⇔⎨ 2 2 ⎩4 cos x = 4 − 12 sin x + 9 sin x 2 ⎧ ⎪sin x ≤ 3 ⇔⎨ ⎪4 1 − sin 2 x = 4 − 12 sin x + 9 sin 2 x ⎩

(

)

⇔

⇔ ⇔ ⇔

2 ⎧ ⎪sin x ≤ 3 ⎨ ⎪13 sin2 x − 12 sin x = 0 ⎩ 2 ⎧ ⎪sin x ≤ 3 ⎪ ⎨ ⎪sin x = 0 ∨ sin x = 12 ⎪ 13 ⎩ sin x = 0 x = kπ, k ∈

Baø i 149 : Giaû i phöông trình sin x cos x + sin x + cos x = 1 ( * )

π⎞ ⎛ Ñaët t = sin x + cos x = 2 sin ⎜ x + ⎟ 4⎠ ⎝
Vôù i ñieà u kieä n : 0 ≤ t ≤ 2 Thì t 2 = 1 + 2sin x cos x t2 − 1 +t =1 Do ñoù (*) thaø n h : 2 ⇔ t 2 + 2t − 3 = 0

⇔ t = 1 ∨ t = −3 ( loaïi )
Vaä y ( * ) ⇔ 12 = 1 + 2sin x cos x ⇔ sin 2x = 0

kπ ,k ∈ 2 Baø i 150 : Giaû i phöông trình ⇔x=

Ñaët t = sin x − cos x ñieàu kieän 0 ≤ t ≤ 2 Thì t = 1 − sin 2x ( *) thaønh : t + 2 1 − t 2 = 1
2

(

sin x − cos x + 2 sin 2x = 1 ( * )

)

(

)

⇔ 2t 2 − t − 1 = 0 1 ⇔ t = 1 ∨ t = − ( loaïi do ñieàu kieän ) 2 2 khi t = 1 thì 1 = 1 − sin 2x

⇔ sin 2x = 0 kπ ⇔x= ,k ∈ 2 Baø i 151 : Giaû i phuông trình sin 4 x − cos4 x = sin x + cos x ( * )

( *) ⇔ ( sin2 x + cos2 x )( sin2 x − cos2 x ) =
⇔ − cos 2x = sin x + cos x

sin x + cos x

⎧cos 2x ≤ 0 ⇔⎨ 2 ⇔ cos 2x = −1 ⎩cos 2x = 1 π ⇔ x = + kπ, k ∈ 2

⎧− cos 2x ≥ 0 ⎪ ⇔⎨ 2 ⎪cos 2x = 1 + 2 sin x cos x ⎩ ⎧cos 2x ≤ 0 ⎪ ⇔⎨ 2 ⎪1 − sin 2x = 1 + sin 2x ⎩ ⎧cos 2x ≤ 0 ⎪ ⇔⎨ 2 ⎪ sin 2x = − sin 2x ⎩ ⎧cos 2x ≤ 0 ⇔⎨ ⎩sin 2x = 0

Baø i 152 : Giaû i phöông trình

3 sin 2x − 2 cos2 x = 2 2 + 2 cos 2x ( *)

Ta coù : ( * ) ⇔ 2 3 sin x cos x − 2 cos2 x = 2 2 + 2 2 cos2 x − 1

(

)

⎛ 3 ⎞ 1 ⇔ cos x ⎜ sin x − cos x ⎟ = cos x ⎜ 2 ⎟ 2 ⎝ ⎠ π⎞ ⎛ ⇔ cos x.sin ⎜ x − ⎟ = cos x 6⎠ ⎝ ⎧cos x > 0 ⎧cos x < 0 ⎪ ⎪ ⇔ cos x = 0 ∨ ⎨ ∨⎨ π⎞ π⎞ ⎛ ⎛ ⎪sin ⎜ x − 6 ⎟ = 1 ⎪sin ⎜ x − 6 ⎟ = −1 ⎝ ⎠ ⎝ ⎠ ⎩ ⎩ ⎧cos x > 0 ⎧cos x < 0 ⎪ ⎪ ⇔ cos x = 0 ∨ ⎨ ∨⎨ π π π π ⎪ x − 6 = 2 + k2π, k ∈ ⎪ x − 6 = − 2 + k2π, k ∈ ⎩ ⎩ ⎧cos x > 0 ⎧cos x < 0 π ⎪ ⎪ ⇔ x = + kπ, k ∈ ∨ ⎨ ∨⎨ 2π π 2 ⎪ x = 3 + k2π, k ∈ ⎪ x = − 3 + k2π, k ∈ ⎩ ⎩ π ⇔ x = + kπ, k ∈ 2

Baø i 153 : Tìm caù c nghieä m treâ n ( 0, 2π ) cuû a phöông trình :

sin 3x − sin x = sin 2x + cos 2x ( *) 1 − cos 2x 2 cos 2x sin x π⎞ ⎛ Ta coù : ( * ) ⇔ = 2 cos ⎜ 2x − ⎟ 4⎠ 2 sin x ⎝
Ñieà u kieä n : sin x ≠ 0 ⇔ x ≠ kπ • Khi x ∈ ( 0, π ) thì sin x > 0 neân :

π⎞ ⎛ 2 cos 2x = 2 cos ⎜ 2x − ⎟ 4⎠ ⎝ π⎞ ⎛ ⇔ 2x = ± ⎜ 2x − ⎟ + k2π, k ∈ 4⎠ ⎝ π ⇔ 4x = + k2π, k ∈ 4 π kπ ⇔x= + ,k ∈ 16 2 π 9π Do x ∈ ( 0, π ) neân x = hay x = 16 16 Khi x ∈ ( π, 2π ) thì sinx < 0 neâ n :

( *) ⇔

⎝ π⎞ ⎛ ⇔ cos ( π − 2x ) = cos ⎜ 2x − ⎟ 4⎠ ⎝ π ⇔ 2x − = ± ( π − 2x ) + k2π, k ∈ 4 5π ⇔ 4x = + k2π, k ∈ 4 5π kπ ⇔x= + ,k ∈ 16 2 21π 29π ∨x= • Do x ∈ ( π, 2π ) neân x = 16 16

( *) ⇔ − cos 2x = cos ⎛ 2x − ⎜

π⎞ ⎟ 4⎠

Baø i 154

Cho phöông trình : sin 6 x + cos6 x = a sin 2x (*) Tìm a sao cho phöông trình coù nghieä m .

Ta coù :

sin6 x + cos6 x = ( sin2 x + cos2 x )( sin4 x − sin2 x cos2 x + cos4 x )

= ( sin2 x + cos2 x ) − 3 sin2 x cos2 x
2

3 sin 2 2x 4 Ñaët t = sin 2x ñieà u kieä n 0 ≤ t ≤ 1

=1−

3 2 t = at ( * *) 4 1 3 ⇔ − t = a (do t = 0 thì (**) voâ nghieä m ) t 4 1 3 Xeù t y = − t treân D = ( 0,1] t 4 1 3 thì y ' = − 2 − < 0 t 4
thì (*) thaø n h : 1 −

Do ñoù : (*) coù nghieä m ⇔ a ≥ Baø i 155 Cho phöông trình

1 • 4

cos 2x = m cos2 x 1 + tgx

( *)

⎡ π⎤ Tìm m ñeå phöông trình coù nghieä m treâ n ⎢0, ⎥ ⎣ 3⎦
Ñaë t t = tgx thì Vaä y : (*) thaø n h: 1 − t 2 = m 1 + t ( * *) (chia 2 veá cho cos2 ≠ 0 )

π thì t ∈ ⎡0, 3 ⎤ ⎣ ⎦ 3 (1 − t )(1 + t ) = 1 − t 1 + t 1 − t2 = Vaä y (**) ⇔ m = ( ) 1+ t 1+ t Xeù t y = (1 − t ) 1 + t treân ⎡0, 3 ⎤ ⎣ ⎦
Khi 0 ≤ x ≤ Ta coù
y' = − 1+ t +

(1 − t )

2 1+ t 2 1+t −3t − 1 ⇔ y' = < 0 ∀t ∈ ⎡0, 3 ⎤ ⎣ ⎦ 2 1+t

=

−2 (1 + t ) + (1 − t )

⎡ π⎤ Do ñoù : (*) coù nghieä m treâ n ⎢0, ⎥ ⇔ 1 − 3 ⎣ 3⎦

(

)

1+ 3 ≤ m ≤ 1•

BAØI TAÄP
1. Giaû i caù c phöông trình a/ sin x − cox = 1 − 4 sin 2x

b/ 4 sin x + 3 cos x = 3 c/ tgx = cot gx + 1 cos x

⎛ 1 + 3 cos2 x ⎞ 1 1 1 d/ + − 2 = − 2⎜ ⎟ 2 sin x 1 − cos x 1 + cos x ⎝ sin x ⎠ 1 e/ cot gx = tgx + sin x f/ 2 cos x − sin x = 1
g/ h/ m/ 1 + cos x + 1 − cos x = 4 sin x cos x 1 − cos 2x 1⎞ ⎛ = 2 ⎜ cos x − ⎟ sin x 2⎠ ⎝ cos 2x + 1 + sin 2x = sin 3 x + cos3 x 2

n/ cos x + sin 3x = 0 1 sin x s/ cos x + 2 sin 2x − cos 3x = 1 + 2 sin x − cos 2x r/ cot gx = tgx + tg 2 x 1 = tgx + 1 + o/ tgx − 1 tgx − 1 p/ sin x − cos x + sin x + cos x = 2
2. sin x + cos x + a sin 2x = 1 Tìm tham soá a döông sao cho phöông trình coù nghieä m 3. Cho phöông trình: sin x − cos x + 4 sin 2x = m a/ Giaû i phöông trình khi m = 0 b/ Tìm m ñeå phöông trình coù nghieä m (ÑS

2−4≤m≤

65 ) 16

Th.S Phạm Hồng Danh (TT luyện thi ĐH Vĩnh Viễn)

```
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# 20588537-Chuong7-Phuong-Trinh-Luong-Giac-Chua-Can-Thuc-Va-Gia-Tri-Tuyet-Doi

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```							CHÖÔNG VII

PHÖÔNG TRÌNH LÖÔÏ N G GIAÙ C CHÖÙ A CAÊ N VAØ PHÖÔNG TRÌNH LÖÔÏ N G GIAÙ C CHÖÙ A GIAÙ TRÒ TUYEÄT ÑOÁI
A) PHÖÔNG TRÌNH LÖÔÏ N G GIAÙ C CHÖÙ A CAÊ N Caù c h giaû i : AÙ p duï n g caù c coâ n g thöù c ⎧A ≥ 0 A = B⇔⎨ ⇔ ⎩A = B

⎧B ≥ 0 A =B⇔⎨ 2 ⎩A = B Ghi chuù : Do theo phöông trình chænh lyù ñaõ boû phaà n baá t phöông trình löôï n g giaù c neâ n ta xöû lyù ñieà u kieä n B ≥ 0 baè n g phöông phaù p thöû laï i vaø chuù n g toâ i boû caù c baø i toaù n quaù phöù c taï p .

⎧B ≥ 0 ⎨ ⎩A = B

Baø i 138 : Giaû i phöông trình

5 cos x − cos 2x + 2 sin x = 0 ( *)

( *) ⇔

5 cos x − cos 2x = −2 sin x

⎧sin x ≤ 0 ⇔⎨ 2 ⎩5 cos x − cos 2x = 4 sin x ⎧sin x ≤ 0 ⎪ ⇔⎨ 2 2 ⎪5 cos x − 2 cos x − 1 = 4 1 − cos x ⎩ ⎧sin x ≤ 0 ⇔⎨ 2 ⎩2 cos x + 5 cos x − 3 = 0 ⎧sin x ≤ 0 ⎪ ⇔⎨ 1 ⎪cos x = 2 ∨ cos x = −3 ( loaïi ) ⎩ ⎧sin x ≤ 0 ⎪ ⇔⎨ π ⎪ x = ± 3 + k2π, k ∈ ⎩ π ⇔ x = − + k2π, k ∈ 3

(

)

(

)

Baø i 139 : Giaû i phöông trình sin3 x + cos3 x + sin3 x cot gx + cos3 xtgx = 2 sin 2x

Ñieà u kieä n : ⎧cos x ≠ 0 ⎪ ⎨sin x ≠ 0 ⇔ ⎪sin 2x ≥ 0 ⎩

⎧sin 2x ≠ 0 ⇔ sin 2x > 0 ⎨ ⎩sin 2x ≥ 0

Luù c ñoù : ( *) ⇔ sin3 x + cos3 x + sin2 x cos x + cos2 x sin x = 2 sin 2x

⇔ sin2 x ( sin x + cos x ) + cos2 x ( cos x + sin x ) = 2sin 2x
⇔ ( sin x + cos x ) sin 2 x + cos2 x = 2 sin 2x

(

)

⎧sin x + cos x ≥ 0 ⎪ ⇔⎨ 2 ⎪( sin x + cos x ) = 2 sin 2x ⎩
⎧ π⎞ ⎛ ⎧ π⎞ ⎛ ⎪ 2 sin ⎜ x + ⎟ ≥ 0 ⎪sin ⎜ x + ⎟ ≥ 0 4⎠ ⇔⎨ ⇔⎨ 4⎠ ⎝ ⎝ ⎪1 + sin 2x = 2 sin 2x ⎪sin 2x = 1 ( nhaän do sin 2x > 0 ) ⎩ ⎩ ⎧ π⎞ ⎧ π⎞ ⎛ ⎛ ⎪sin ⎜ x + 4 ⎟ ≥ 0 ⎪sin ⎜ x + 4 ⎟ ≥ 0 ⎪ ⎪ ⎝ ⎠ ⎝ ⎠ ⇔⎨ ⇔⎨ ⎪ x = π + kπ, k ∈ ⎪ x = π + m2π ∨ x = 5π + m2π ( loaïi ) , m ∈ ⎪ ⎪ ⎩ 4 ⎩ 4 4 π ⇔ x = + m2π, m ∈ 4

Baø i 140 : Giaû i phöông trình

π⎞ ⎛ 1 + 8 sin 2x. cos2 2x = 2 sin ⎜ 3x + ⎟ ( *) 4⎠ ⎝

⎧ π⎞ ⎛ ⎪sin ⎜ 3x + 4 ⎟ ≥ 0 ⎪ ⎝ ⎠ Ta coù : (*) ⇔ ⎨ ⎪1 + 8 sin 2x cos2 2x = 4 sin2 ⎛ 3x + π ⎞ ⎜ ⎟ ⎪ 4⎠ ⎝ ⎩ ⎧ π⎞ ⎛ ⎪sin ⎜ 3x + 4 ⎟ ≥ 0 ⎪ ⎝ ⎠ ⇔⎨ ⎪1 + 4 sin 2x (1 + cos 4x ) = 2 ⎡1 − cos( 6x + π ) ⎤ ⎢ ⎪ 2 ⎥ ⎣ ⎦ ⎩ ⎧ π⎞ ⎛ ⎪sin ⎜ 3x + ⎟ ≥ 0 4⎠ ⇔⎨ ⎝ ⎪1 + 4 sin 2x + 2 ( sin 6x − sin 2x ) = 2 (1 + sin 6x ) ⎩
⎧ π⎞ ⎧ π⎞ ⎛ ⎛ ⎪sin ⎜ 3x + 4 ⎟ ≥ 0 ⎪sin ⎜ 3x + 4 ⎟ ≥ 0 ⎪ ⎪ ⎝ ⎠ ⎝ ⎠ ⇔⎨ ⇔⎨ ⎪sin 2x = 1 ⎪ x = π + kπ ∨ x = 5π + kπ, k ∈ ⎪ ⎪ ⎩ 2 ⎩ 12 12

π⎞ ⎛ So laï i vôù i ñieà u kieä n sin ⎜ 3x + ⎟ ≥ 0 4⎠ ⎝ π •Khi x = + kπ thì 12 π⎞ ⎛ ⎛π ⎞ sin ⎜ 3x + ⎟ = sin ⎜ + 3kπ ⎟ = cos kπ 4⎠ ⎝ ⎝2 ⎠ ⎡1 , ( neáu k chaün ) ( nhaän ) =⎢ ⎢ −1 , ( neáu k leû ) ( loaïi ) ⎣ 5π • Khi x = + kπ thì 12 π⎞ ⎛ ⎛ 3π ⎞ ⎛ π ⎞ sin ⎜ 3x + ⎟ = sin ⎜ + 3kπ ⎟ = sin ⎜ − + kπ ⎟ 4⎠ ⎝ ⎝ 2 ⎠ ⎝ 2 ⎠

⎡ −1 , neáu k chaün ( loaïi ) =⎢ ⎢1 , neáu k leû ( nhaän ) ⎣ π 5π + m2π ∨ x = + ( 2m + 1) π, m ∈ Do ñoù ( *) ⇔ x = 12 12
1 − sin 2x + 1 + sin 2x = 4 cos x ( * ) sin x

Baø i 141 : Giaû i phöông trình

Luù c ñoù : ( *) ⇔ 1 − sin 2x + 1 + sin 2x = 2 sin 2x ( hieå n nhieâ n sinx = 0 khoâ n g laø nghieä m , vì sinx =0 thì VT = 2, VP = 0 ) ⎧ ⎪2 + 2 1 − sin2 2x = 4 sin2 2x ⇔⎨ ⎪sin 2x ≥ 0 ⎩ ⎧ 1 − sin2 2x = 2 sin2 2x − 1 ⎪ ⇔⎨ ⎪sin 2x ≥ 0 ⎩

⇔

⇔

⇔

⇔

⎧1 − sin 2 2x = 4 sin4 2x − 4 sin2 2x + 1 ⎪ 1 ⎪ 2 ⎨sin 2x ≥ 2 ⎪ ⎪sin 2x ≥ 0 ⎩ ⎧sin 2 2x 4 sin 2 2x − 3 = 0 ⎪ ⎨ 1 ⎪sin 2x ≥ 2 ⎩ ⎧ 3 − 3 ∨ sin 2x = ⎪sin 2x = ⎪ 2 2 ⎨ ⎪sin 2x ≥ 2 ⎪ 2 ⎩ 3 sin 2x = 2

(

)

π 2π + k2π ∨ 2x = + k2π, k ∈ 3 3 π π ⇔ x = + kπ ∨ x = + kπ, k ∈ 6 3 Chuù yù : Coù theå ñöa veà phöông trình chöù a giaù trò tuyeä t ñoá i ⎧sin x ≠ 0 ( *) ⇔ ⎪ ⎨ ⎪ cos x − sin x + cos x + sin x = 2 sin 2x ⎩ ⇔ cos x − sin x + cos x + sin x = 2 sin 2x ⇔ 2x =
Baø i 142 : Giaû i phöông trình sin x + 3 cos x + sin x + 3 cos x = 2 ( * )
π 3 cos x Ñaët t = sin x + 3 cos x = sin x + π cos 3 π⎞ π⎞ 1 ⎛ ⎛ ⇔t= sin ⎜ x + ⎟ = 2 sin ⎜ x + ⎟ π 3⎠ 3⎠ ⎝ ⎝ cos 3 ( *) thaønh t + t = 2 sin

⇔

t = 2−t

Do ñoù ( * )

⎧2 − t ≥ 0 ⎧t ≤ 2 ⇔⎨ ⇔⎨ 2 2 ⎩t = 4 − 4t + t ⎩t − 5t + 4 = 0 ⎧t ≤ 2 ⇔⎨ ⇔ t =1 ⎩t = 1 ∨ t = 4

π⎞ 1 π π π 5π ⎛ ⇔ sin ⎜ x + ⎟ = ⇔ x + = + k2π hay x + = + k2π, k ∈ 3⎠ 2 3 6 3 6 ⎝ π π ⇔ x = − + k2π ∨ x = + k2π, k ∈ 6 2

Baø i 143 : Giaû i phöông trình 3 tgx + 1 ( sin x + 2 cos x ) = 5 ( sin x + 3 cos x ) ( *) Chia hai veá cuû a (*) cho cos x ≠ 0 ta ñöôï c ( *) ⇔ 3 tgx + 1 ( tgx + 2) = 5 ( tgx + 3) Ñaët u =
tgx + 1 vôùi u ≥ 0

Thì u 2 − 1 = tgx

(*) thaøn h 3u u 2 + 1 = 5 u 2 + 2

(

)

(

)

⇔ 3u 3 − 5u 2 + 3u − 10 = 0 ⇔ ( u − 2 ) ( 3u 2 + u + 5 ) = 0
⇔ u = 2 ∨ 3u 2 + u + 5 = 0 ( voâ nghieäm )

Do ñoù

( *) ⇔

tgx + 1 = 2

⇔ tgx + 1 = 4
π π⎞ ⎛ ⇔ tgx = 3 = tgα ⎜ vôùi − < α < ⎟ ⇔ x = α + k π , k ∈ 2 2⎠ ⎝

Baø i 144 : Giaû i phöông trình

(

1 − cos x + cos x cos 2x =

)

( *) ⇔ (

1 − cos x + cos x cos 2x = sin 2x cos 2x

)

1 sin 4x ( *) 2

⎧cos x ≥ 0 ⇔⎨ hay ⎩cos 2x = 0
⎧cos x ≥ 0 ⎪ ⇔⎨ π ⎪2x = 2 + kπ, k ∈ ⎩

1 − cos x + cos x = sin 2x

⎧cos x ≥ 0 ⎪ hay ⎨sin 2x ≥ 0 ⎪ 2 ⎩1 + 2 ( 1 − cos x)cosx = sin 2x ⎧cos x ≥ 0 ⎧cos x ≥ 0 ⎪ ⎪ hay ⎨sin 2x ≥ 0 ⇔⎨ π π ⎪x = 4 + k 2 , k ∈ ⎪ 2 ⎩ ⎩1 + 2 ( 1 − cos x)cosx = sin 2x ( VT ≥ 1 ≥ VP ) ⎧cos x ≥ 0 cos x ≥ 0 ⎪sin 2x ≥ 0 ⎧ ⎪ ⎪ ⇔⎨ hay ⎨ 2 π 5π ⎪ x = ± 4 + hπ hay x = ± 4 + hπ, h ∈ ⎪sin 2x = 1 ⎩ ⎪ ⎩(1 − cos x ) cos x = 0 π ⇔ x = ± + hπ, h ∈ 4 ⎧sin 2x = 1 ⎧sin 2x = 1 hay ⎨ hay ⎨ ⎩cos x = 0 ( ⇒ sin 2x = 0 ) ⎩cos x = 1 ( ⇒ sin x = 0 ⇒ sin 2x = 0 ) π ⇔ x = ± + hπ, h ∈ 4

Baø i 145 : Giaû i phöông trình sin3 x (1 + cot gx ) + cos3 x (1 + tgx ) = 2 sin x cos x ( *)

( *) ⇔ sin3 x ⎛ ⎜

⇔ ( sin x + cos x ) sin 2 x + cos2 x = 2 sin x cos x

sin x + cos x ⎞ ⎛ cos x + sin x ⎞ 3 ⎟ + cos x ⎜ ⎟ = 2 sin x cos x sin x cos x ⎝ ⎠ ⎝ ⎠

(

)

⎧sin x + cos x ≥ 0 ⇔⎨ ⎩1 + sin 2x = 2 sin 2x
⎧ π⎞ ⎛ ⎪sin ⎜ x + 4 ⎟ ≥ 0 ⎧sin x + cos x ≥ 0 ⎪ ⎝ ⎠ ⇔⎨ ⇔⎨ ⎩sin 2x = 1 ⎪ x = π + kπ, k ∈ ⎪ 4 ⎩

⎧ π⎞ ⎛ ⎪sin ⎜ x + 4 ⎟ ≥ 0 ⎪ ⎝ ⎠ ⇔⎨ ⎪ x + π = π + kπ, k ∈ ⎪ 4 2 ⎩ ⎧ π⎞ ⎛ ⎪sin ⎜ x + 4 ⎟ ≥ 0 ⎪ ⎝ ⎠ ⇔⎨ ⎪ x + π = π + h2π hay x + π = 3π + h2π, h ∈ ⎪ ⎩ 4 2 4 2 π ⇔ x = + h2π, h ∈ 4

Baø i 146 : Giaû i phöông trình

cos 2x + 1 + sin 2x = 2 sin x + cos x ( *)

π⎞ ⎛ Ñieà u kieä n cos 2x ≥ 0 vaø sin ⎜ x + ⎟ ≥ 0 4⎠ ⎝

Luù c ñoù : ( *) ⇔

cos2 x − sin 2 x +
2

( cos x + sin x )

2

= 2 cos x + sin x
2

⇔ cos2 x − sin 2 x + ( cos x + sin x ) + 2 cos 2x

⇔ cos x ( cos x + sin x ) + ( sin x + cos x ) cos 2x

( cos x + sin x ) = 4 ( sin x + cos x ) = 2 ( sin x + cos x )

⎡sin x + cos x = 0 ⇔⎢ ⎣cos x + cos 2x = 2 ⎡ tgx = −1 ⇔⎢ ⎢ cos 2x = 2 − cos x ( * *) ⎣ ⎡ tgx = −1 ⇔⎢ 2 ⎣cos 2x = 4 − 4 cos x + cos x ⇔ tgx = −1 ∨ cos2 x + 4 cos x − 5 = 0 ⇔ tgx = −1 ∨ cos x = 1 ∨ cos x = −5 ( loaïi )

π + kπ ∨ x = k2π, k ∈ 4 π ⎛ π⎞ Thöû laï i : • x = − + kπ thì cos 2x = cos ⎜ − ⎟ = 0 ( nhaän ) 4 ⎝ 2⎠ π⎞ ⎛ Vaø sin ⎜ x + ⎟ = sin kπ = 0 ( nhaän ) 4⎠ ⎝ • x = k2π thì cos 2x = 1 ( nhaän ) ⇔x=−
π⎞ π ⎛ vaø cos ⎜ x + ⎟ = cos > 0 ( nhaän ) 4⎠ 4 ⎝ π Do ñoù (*) ⇔ x = − + kπ ∨ x = k2π, k ∈ 4 Chuù yù : Taï i (**) coù theå duø n g phöông trình löôï n g giaù c khoâ n g möï c

( * *) ⇔ ⎪ ⎨

⎧cos x = 1 ⇔⎨ ⇔ x = 2kπ, k ∈ ⎩sin x + cos x ≥ 0 Caù c h khaù c

⎧cos x = 1 ⎪ ⇔ ⎨cos 2x = 2 cos2 x − 1 = 1 ⎪sin x + cos x ≥ 0 ⎩

⎧cos x + cos 2x = 2 ⎪sin x + cos x ≥ 0 ⎩

( *) ⇔
⇔

cos2 x − sin 2 x +

( cos x + sin x )

2

= 2 cos x + sin x
2

(cos x + sin x).(cos x − sin x ) +

( cos x + sin x )

= 2 cos x + sin x

⎧cos x + sin x > 0 ⎪ ⇔ cos x + sin x = 0 hay ⎨ ⎪ cos x − sin x + ⎩ ⎧cos x + sin x > 0 ⎪ ⇔ tgx = − 1 hay ⎨ ⎪2 cos x + 2 cos 2x = 4 ⎩

( cos x + sin x ) = 2

⎧cos x + sin x > 0 ⎪ ⇔ tgx = − 1 hay ⎨ ⎪cos x + cos 2x = 2 ⎩ ⎧cos x = 1 π ⇔ x = − + kπ, k ∈ hay ⎨ 4 ⎩cos 2x = 1 π ⇔ x = − + kπ hay x = 2kπ, k ∈ 4 ( nhaä n xeù t : khi cosx =1 thì sinx = 0 vaø sinx + cosx = 1 > 0 )

1. Giaû i phöông trình : a/ 1 + sin x + cos x = 0 4x − cos2 x cos 3 =0 b/ 1 − tg 2 x d/

BAØI TAÄP

c/ sin x + 3 cos x = 2 + cos 2x + 3 sin 2x
sin 2 x − 2 sin x + 2 = 2 sin x − 1 3tgx e/ 2 3 sin x = − 3 2 sin x − 1

sin2 2x + cos4 2x − 1 =0 f/ sin cos x g/ 8 cos 4x cos2 2x + 1 − cos 3x + 1 = 0
h/

sin x + sin x + sin2 x + cos x = 1

k/ 5 − 3sin 2 x − 4 cos x = 1 − 2 cos x l/ cos 2x = cos2 x 1 + tgx 2. Cho phöông trình : 1 + sin x + 1 − sin x = m cos x (1) a/ Giaû i phöông trình khi m = 2 b/ Giaû i vaø bieä n luaä n theo m phöông trình (1) 3. Cho f(x) = 3cos 6 2x + sin 42x + cos4x – m a/ Giaû i phöông trình f(x) = 0 khi m = 0 b/ Cho g ( x ) = 2 cos2 2x 3 cos2 2x + 1 . Tìm taá t caû caù c giaù trò m ñeå phöông trình f(x) = g(x) coù nghieä m . ( ÑS : 1 ≤ m ≤ 0 ) 4. Tìm m ñeå phöông trình sau coù nghieä m 1 + 2 cos x + 1 + 2sin x = m

(ÑS :

1+ 3 ≤ m ≤ 2 1+ 2

)

B) PHÖÔNG TRÌNH LÖÔÏ N G GIAÙ C CHÖÙ A CAÙ C TRÒ TUYEÄ T ÑOÁ I Caù ch giaû i : 1/ Môû giaù trò tuyeä t ñoá i baè n g ñònh nghóa 2/ AÙ p duï n g • A = B ⇔ A = ±B

⎧B ≥ 0 ⎧B ≥ 0 ⎧A ≥ 0 ⎧A < 0 •A =B⇔⎨ ⇔⎨ 2 ⇔⎨ ∨⎨ 2 ⎩ A = ±B ⎩ A = B ⎩ A = −B ⎩A = B
Baø i 147 : Giaû i phöông trình cos 3x = 1 − 3 sin 3x ( *)

( *) ⇔ ⎪ ⎨

⎧1 − 3 sin 3x ≥ 0

1 ⎧ ⎪sin 3x ≤ 3 ⇔⎨ ⎪4 sin 2 3x − 2 3 sin 3x = 0 ⎩

2 2 ⎪cos 3x = 1 − 2 3 sin 3x + 3sin 3x ⎩ 1 ⎧ ⎪sin 3x ≤ 3 ⇔⎨ ⎪1 − sin 2 3x = 1 − 2 3 sin 3x + 3 sin 2 3x ⎩

1 ⎧ ⎪sin 3x ≤ 3 ⎪ ⇔⎨ ⎪sin 3x = 0 ∨ sin 3x = 3 ⎪ ⎩ 2 ⇔ sin 3x = 0

⇔x=

kπ ,k ∈ 3

Baø i 148 : Giaû i phöông trình 3sin x + 2 cos x − 2 = 0 ( * )

( *) ⇔ 2 cos x

= 2 − 3sin x

⎧2 − 3sin x ≥ 0 ⇔⎨ 2 2 ⎩4 cos x = 4 − 12 sin x + 9 sin x 2 ⎧ ⎪sin x ≤ 3 ⇔⎨ ⎪4 1 − sin 2 x = 4 − 12 sin x + 9 sin 2 x ⎩

(

)

⇔

⇔ ⇔ ⇔

2 ⎧ ⎪sin x ≤ 3 ⎨ ⎪13 sin2 x − 12 sin x = 0 ⎩ 2 ⎧ ⎪sin x ≤ 3 ⎪ ⎨ ⎪sin x = 0 ∨ sin x = 12 ⎪ 13 ⎩ sin x = 0 x = kπ, k ∈

Baø i 149 : Giaû i phöông trình sin x cos x + sin x + cos x = 1 ( * )

π⎞ ⎛ Ñaët t = sin x + cos x = 2 sin ⎜ x + ⎟ 4⎠ ⎝
Vôù i ñieà u kieä n : 0 ≤ t ≤ 2 Thì t 2 = 1 + 2sin x cos x t2 − 1 +t =1 Do ñoù (*) thaø n h : 2 ⇔ t 2 + 2t − 3 = 0

⇔ t = 1 ∨ t = −3 ( loaïi )
Vaä y ( * ) ⇔ 12 = 1 + 2sin x cos x ⇔ sin 2x = 0

kπ ,k ∈ 2 Baø i 150 : Giaû i phöông trình ⇔x=

Ñaët t = sin x − cos x ñieàu kieän 0 ≤ t ≤ 2 Thì t = 1 − sin 2x ( *) thaønh : t + 2 1 − t 2 = 1
2

(

sin x − cos x + 2 sin 2x = 1 ( * )

)

(

)

⇔ 2t 2 − t − 1 = 0 1 ⇔ t = 1 ∨ t = − ( loaïi do ñieàu kieän ) 2 2 khi t = 1 thì 1 = 1 − sin 2x

⇔ sin 2x = 0 kπ ⇔x= ,k ∈ 2 Baø i 151 : Giaû i phuông trình sin 4 x − cos4 x = sin x + cos x ( * )

( *) ⇔ ( sin2 x + cos2 x )( sin2 x − cos2 x ) =
⇔ − cos 2x = sin x + cos x

sin x + cos x

⎧cos 2x ≤ 0 ⇔⎨ 2 ⇔ cos 2x = −1 ⎩cos 2x = 1 π ⇔ x = + kπ, k ∈ 2

⎧− cos 2x ≥ 0 ⎪ ⇔⎨ 2 ⎪cos 2x = 1 + 2 sin x cos x ⎩ ⎧cos 2x ≤ 0 ⎪ ⇔⎨ 2 ⎪1 − sin 2x = 1 + sin 2x ⎩ ⎧cos 2x ≤ 0 ⎪ ⇔⎨ 2 ⎪ sin 2x = − sin 2x ⎩ ⎧cos 2x ≤ 0 ⇔⎨ ⎩sin 2x = 0

Baø i 152 : Giaû i phöông trình

3 sin 2x − 2 cos2 x = 2 2 + 2 cos 2x ( *)

Ta coù : ( * ) ⇔ 2 3 sin x cos x − 2 cos2 x = 2 2 + 2 2 cos2 x − 1

(

)

⎛ 3 ⎞ 1 ⇔ cos x ⎜ sin x − cos x ⎟ = cos x ⎜ 2 ⎟ 2 ⎝ ⎠ π⎞ ⎛ ⇔ cos x.sin ⎜ x − ⎟ = cos x 6⎠ ⎝ ⎧cos x > 0 ⎧cos x < 0 ⎪ ⎪ ⇔ cos x = 0 ∨ ⎨ ∨⎨ π⎞ π⎞ ⎛ ⎛ ⎪sin ⎜ x − 6 ⎟ = 1 ⎪sin ⎜ x − 6 ⎟ = −1 ⎝ ⎠ ⎝ ⎠ ⎩ ⎩ ⎧cos x > 0 ⎧cos x < 0 ⎪ ⎪ ⇔ cos x = 0 ∨ ⎨ ∨⎨ π π π π ⎪ x − 6 = 2 + k2π, k ∈ ⎪ x − 6 = − 2 + k2π, k ∈ ⎩ ⎩ ⎧cos x > 0 ⎧cos x < 0 π ⎪ ⎪ ⇔ x = + kπ, k ∈ ∨ ⎨ ∨⎨ 2π π 2 ⎪ x = 3 + k2π, k ∈ ⎪ x = − 3 + k2π, k ∈ ⎩ ⎩ π ⇔ x = + kπ, k ∈ 2

Baø i 153 : Tìm caù c nghieä m treâ n ( 0, 2π ) cuû a phöông trình :

sin 3x − sin x = sin 2x + cos 2x ( *) 1 − cos 2x 2 cos 2x sin x π⎞ ⎛ Ta coù : ( * ) ⇔ = 2 cos ⎜ 2x − ⎟ 4⎠ 2 sin x ⎝
Ñieà u kieä n : sin x ≠ 0 ⇔ x ≠ kπ • Khi x ∈ ( 0, π ) thì sin x > 0 neân :

π⎞ ⎛ 2 cos 2x = 2 cos ⎜ 2x − ⎟ 4⎠ ⎝ π⎞ ⎛ ⇔ 2x = ± ⎜ 2x − ⎟ + k2π, k ∈ 4⎠ ⎝ π ⇔ 4x = + k2π, k ∈ 4 π kπ ⇔x= + ,k ∈ 16 2 π 9π Do x ∈ ( 0, π ) neân x = hay x = 16 16 Khi x ∈ ( π, 2π ) thì sinx < 0 neâ n :

( *) ⇔

⎝ π⎞ ⎛ ⇔ cos ( π − 2x ) = cos ⎜ 2x − ⎟ 4⎠ ⎝ π ⇔ 2x − = ± ( π − 2x ) + k2π, k ∈ 4 5π ⇔ 4x = + k2π, k ∈ 4 5π kπ ⇔x= + ,k ∈ 16 2 21π 29π ∨x= • Do x ∈ ( π, 2π ) neân x = 16 16

( *) ⇔ − cos 2x = cos ⎛ 2x − ⎜

π⎞ ⎟ 4⎠

Baø i 154

Cho phöông trình : sin 6 x + cos6 x = a sin 2x (*) Tìm a sao cho phöông trình coù nghieä m .

Ta coù :

sin6 x + cos6 x = ( sin2 x + cos2 x )( sin4 x − sin2 x cos2 x + cos4 x )

= ( sin2 x + cos2 x ) − 3 sin2 x cos2 x
2

3 sin 2 2x 4 Ñaët t = sin 2x ñieà u kieä n 0 ≤ t ≤ 1

=1−

3 2 t = at ( * *) 4 1 3 ⇔ − t = a (do t = 0 thì (**) voâ nghieä m ) t 4 1 3 Xeù t y = − t treân D = ( 0,1] t 4 1 3 thì y ' = − 2 − < 0 t 4
thì (*) thaø n h : 1 −

Do ñoù : (*) coù nghieä m ⇔ a ≥ Baø i 155 Cho phöông trình

1 • 4

cos 2x = m cos2 x 1 + tgx

( *)

⎡ π⎤ Tìm m ñeå phöông trình coù nghieä m treâ n ⎢0, ⎥ ⎣ 3⎦
Ñaë t t = tgx thì Vaä y : (*) thaø n h: 1 − t 2 = m 1 + t ( * *) (chia 2 veá cho cos2 ≠ 0 )

π thì t ∈ ⎡0, 3 ⎤ ⎣ ⎦ 3 (1 − t )(1 + t ) = 1 − t 1 + t 1 − t2 = Vaä y (**) ⇔ m = ( ) 1+ t 1+ t Xeù t y = (1 − t ) 1 + t treân ⎡0, 3 ⎤ ⎣ ⎦
Khi 0 ≤ x ≤ Ta coù
y' = − 1+ t +

(1 − t )

2 1+ t 2 1+t −3t − 1 ⇔ y' = < 0 ∀t ∈ ⎡0, 3 ⎤ ⎣ ⎦ 2 1+t

=

−2 (1 + t ) + (1 − t )

⎡ π⎤ Do ñoù : (*) coù nghieä m treâ n ⎢0, ⎥ ⇔ 1 − 3 ⎣ 3⎦

(

)

1+ 3 ≤ m ≤ 1•

BAØI TAÄP
1. Giaû i caù c phöông trình a/ sin x − cox = 1 − 4 sin 2x

b/ 4 sin x + 3 cos x = 3 c/ tgx = cot gx + 1 cos x

⎛ 1 + 3 cos2 x ⎞ 1 1 1 d/ + − 2 = − 2⎜ ⎟ 2 sin x 1 − cos x 1 + cos x ⎝ sin x ⎠ 1 e/ cot gx = tgx + sin x f/ 2 cos x − sin x = 1
g/ h/ m/ 1 + cos x + 1 − cos x = 4 sin x cos x 1 − cos 2x 1⎞ ⎛ = 2 ⎜ cos x − ⎟ sin x 2⎠ ⎝ cos 2x + 1 + sin 2x = sin 3 x + cos3 x 2

n/ cos x + sin 3x = 0 1 sin x s/ cos x + 2 sin 2x − cos 3x = 1 + 2 sin x − cos 2x r/ cot gx = tgx + tg 2 x 1 = tgx + 1 + o/ tgx − 1 tgx − 1 p/ sin x − cos x + sin x + cos x = 2
2. sin x + cos x + a sin 2x = 1 Tìm tham soá a döông sao cho phöông trình coù nghieä m 3. Cho phöông trình: sin x − cos x + 4 sin 2x = m a/ Giaû i phöông trình khi m = 0 b/ Tìm m ñeå phöông trình coù nghieä m (ÑS

2−4≤m≤

65 ) 16

Th.S Phạm Hồng Danh (TT luyện thi ĐH Vĩnh Viễn)

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