How would you tackle $\sin(2x)-2\sin(3x)+\sin(4x)=0$? Just wondering how would you tackle this trig equation.  I tried a lot of different approaches but I end up expanding. Even my prof couldn't answer it earlier. I'm just curious .
$$
\sin(2x)-2\sin(3x)+\sin(4x)=0
$$
 A: $$\sin(2x)+\sin(4x)=2\sin(3x)\cos(x)$$
So
$$\sin(3x) (\cos(x)-1)=0$$
A: using $\sin x+\sin y=2\sin(\frac{x+y}2)\cos(\frac{x-y}2)$ formula we get:
$$\sin3x(\cos x-1)=0$$
$$\implies \sin 3x=0 $$or $$\cos x=1$$
so$$x=n\pi/3$$or$$x=2n\pi$$ where $n\in  \mathbb{Z}$. now it is cakewalk to find final equation.
A: Since $\sin (n x) = \dfrac{\mathrm{e}^{i n x} - \mathrm{e}^{-i n x}}{2 i}$ and $\cos (n x) = \dfrac{\mathrm{e}^{i n x} + \mathrm{e}^{-i n x}}{2}$,
$$\begin{array}{rl} \sin(2x)-2\sin(3x)+\sin(4x) &= \left(\dfrac{\mathrm{e}^{i 2 x} - \mathrm{e}^{-i 2 x}}{2 i}\right) - 2 \left(\dfrac{\mathrm{e}^{i 3 x} - \mathrm{e}^{-i 3 x}}{2 i}\right) + \left(\dfrac{\mathrm{e}^{i 4 x} - \mathrm{e}^{-i 4 x}}{2 i}\right)\\\\ &= \dfrac{1}{2 i}\left(\mathrm{e}^{i 2 x} - 2 \, \mathrm{e}^{i 3 x} + \mathrm{e}^{i 4 x} - \mathrm{e}^{-i 2 x} + 2 \, \mathrm{e}^{-i 3 x} - \mathrm{e}^{-i 4 x}\right)\\\\ &= \dfrac{1}{2 i}\left(\left(\mathrm{e}^{-i x} - 2  + \mathrm{e}^{i x}\right) \mathrm{e}^{i 3 x} - \left(\mathrm{e}^{i x} - 2 + \mathrm{e}^{-i x}\right)\mathrm{e}^{-i 3 x}\right)\\\\ &= \left(\mathrm{e}^{i x} - 2 + \mathrm{e}^{-i x}\right) \sin (3x)\\\\ &= (2\cos(x)-2) \, \sin (3x)\\\\ &= 2(\cos (x) - 1) \sin (3x)\end{array}$$
A: By dividing both sides of our equation by $\sin x$ and letting $t=\cos x$ we get:
$$ U_1(t)-2U_2(t)+U_3(t) = 0 \tag{1}$$
where $U_i$ are Chebyshev polynomials of the second kind. $(1)$ is equivalent to:
$$ 0 = 2 - 2 t - 8 t^2 + 8 t^3 = (2t-2)(2t-1)(2t+1).\tag{2}$$
Now to find the the solutions is straightfoward.
A: A rather systematic approach would be to use FFT together with the convolution theorem to create linear equation system to find and then solve a polynomial in the complex exponential.
Strength could solve ANY of these (to the extent we can solve polynomial equations).
$$\sum_{k\in \mathbb{Z}} s_k\sin(kx) + \sum_{k\in \mathbb{Z}} c_k\cos(kx) = \sum_{k\in \mathbb{Z}} d_k{(e^{2\pi ix})}^k = \left/ t = e^{2\pi i x} \right/ = \sum_{k\in \mathbb{Z}} d_kt^k$$
Weakness requires Fourier transforms so it may be overkill depending on readers knowledge.
A: Even if you don't have the "product-to-sum formulas" handy, the "angle-addition formulas" will get you there fast enough:
$$ \sin(2x) \ + \ \sin(4x) \ - \ 2\sin(3x) \ \ = \ \ \sin(3x - x) \ + \ \sin(3x + x) \ - \ 2\sin(3x) $$
$$ = \ \ [ \sin(3x) \cos x \ - \ \cos(3x) \sin x] \ + \ [ \sin(3x) \cos x \ + \ \cos(3x) \sin x] \ - \ 2 \sin(3x) $$
$$ = \ \  \sin(3x) \cos x  \ + \ \sin(3x) \cos x  \ - \ 2 \sin(3x) \ \ = \ \ \ 2 \sin(3x) \cos x  \ - \ 2 \sin(3x) $$
$$ = \ \ 2 \sin(3x) \ ( \cos x  \ - \ 1 ) \ \ = \ \ 0 \ \ . $$
