TITLE: Solving complex trig functions: $\sin2x + \sin3x = \frac{\sqrt{3}}2$
QUESTION [6 upvotes]: How to solve:
$$\sin(2x) + \sin(3x) = \frac{\sqrt{3}}{2}$$
where $x$ is in $[-\pi,\pi]$?
I have no idea what to do with the $\sin(2x) + \sin(3x)$.
Am I supposed to factorise, differentiate, is there some theory I am to apply?

REPLY [2 votes]: The angle addition formulas
$$\sin(2x)=2\sin x\cos x$$
and
$$\begin{align}
\sin(3x)&=\sin x\cos(2x)+\cos x\sin(2x)\\
&=\sin x(2\cos^2x-1)+2\cos^2x\sin x\\
&=\sin x(4\cos^2x-1)
\end{align}$$
turn the equation $\sin(2x)+\sin(3x)=\sqrt3/2$ into $s(4c^2+2c-1)=\sqrt3/2$, where $s$ and $c$ abbreviate $\sin x$ and $\cos x$.  Squaring both sides and replacing $s^2$ with $1-c^2$ leads to a polynomial expression in $c$:
$$(1-c^2)(4c^2+2c-1)^2={3\over4}$$
It's easy to check that $c=\pm{1\over2}$ satisfies this equation, corresponding to the solutions $x=\pi/3$ and $-2\pi/3$ noted by Tunk-Fey in comments.  (There are, of course, two values of $x$ in $[-\pi,\pi]$ for each value of $c$, but you have to go back to the unsquared equation to get the angle with the correct sign for $s$.)  The polynomial expression in $c$, expanded and factored, is
$$(2c-1)(2c+1)(16c^4+16c^3-16c^2-16c+1)=0$$
The quartic factor has two real roots, both in $[0,1]$, neither of which is the cosine of any nice angle.  I get $c\approx0.0592136551698$ and $c\approx0.983859187765$ for the other roots, with $s\approx-0.99824533209$ and $s\approx0.178944401$ as the corresponding values of $s=\sin x$.  These correspond to $x\approx-0.481140676\pi$ and $x\approx0.0572682233\pi$.
The unpleasantry of the quartic factor makes me wonder about the context of the problem.  Could it have just been asking to find a solution?
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