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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?

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  • $\begingroup$ Is my edit correct? Your text didn't make clear if the number is $\frac{\sqrt{3}}{2}$ or $\sqrt{\frac{3}{2}}$ $\endgroup$
    – egreg
    Apr 21 '14 at 10:36
  • $\begingroup$ the first one :) $\endgroup$
    – confused
    Apr 21 '14 at 10:52
  • $\begingroup$ How about introduce complex numbers, solve the cubic using Cardano's method, and then... Just kidding. $\endgroup$
    – evil999man
    Apr 21 '14 at 11:57
  • $\begingroup$ Try tan(x/2)=t. $\endgroup$
    – evil999man
    Apr 21 '14 at 11:58
  • $\begingroup$ where did then tan come from and the t.? i have no idea where you pulled that from. :/ $\endgroup$
    – confused
    Apr 21 '14 at 11:59
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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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(This is not a complete answer.)

Consider the function $$f(x):=\sin(2x)+\sin(3x)-{\sqrt{3}\over2}\ .$$ Since we can guess the zeros ${\pi\over3}$ and ${4\pi\over 3}$ it makes sense to introduce the function $$g(y):=f\left({\pi\over3}+y\right)\ .$$ Indeed, after some computation one obtains $$g(y)=-\sin y\>\bigl(\cos y+\sqrt{3}\sin y+3\cos^2 y-\sin^2 y\bigr)\ .$$ Unfortunately the second factor cannot be reduced in a simple way.

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