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Solve $\sin 2 x+\cos 2 x=0$
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$\sin 2 x+\cos 2 x=0$
$\sin 2 x=-\cos 2 x$
$\frac{\sin 2 x}{\cos 2 x}=-1$
$\tan 2 x=-1 \longrightarrow 1$
$\pi<x<2 \pi$
$2 \pi<2 x<4 \pi$
$\tan x$ is negative in second and fourth quadrant
$\therefore \tan \frac{3 \pi}{4}=\tan \frac{7 \pi}{4}=\ldots \ldots=\tan (4 n-1) \frac{\pi}{4}=-1$
$n=1,2,3, \ldots$
But $2 \pi<2 x<4 \pi$
$\therefore 2 x=\frac{11 \pi}{4}$ or $2 x=\frac{15 \pi}{4}$
$x=\frac{11 \pi}{8}$ or $\frac{15 \pi}{8}$
by Diamond (38,951 points)

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The text passage contains a trigonometric equation to solve and provides a detailed solution to the problem.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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14 views
Solve $\sin 2 x+\cos 2 x=0$
solved
| 14 views

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$\sin 2 x+\cos 2 x=0$
$\sin 2 x=-\cos 2 x$
$\frac{\sin 2 x}{\cos 2 x}=-1$
$\tan 2 x=-1 \longrightarrow 1$
$\pi<x<2 \pi$
$2 \pi<2 x<4 \pi$
$\tan x$ is negative in second and fourth quadrant
$\therefore \tan \frac{3 \pi}{4}=\tan \frac{7 \pi}{4}=\ldots \ldots=\tan (4 n-1) \frac{\pi}{4}=-1$
$n=1,2,3, \ldots$
But $2 \pi<2 x<4 \pi$
$\therefore 2 x=\frac{11 \pi}{4}$ or $2 x=\frac{15 \pi}{4}$
$x=\frac{11 \pi}{8}$ or $\frac{15 \pi}{8}$
by Diamond (38,951 points)

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The text passage contains a trigonometric equation to solve and provides a detailed solution to the problem.

TYPE: SINGLE_QUESTION_WITH_ANSWER