I know how to solve for $\theta$ using the general solution method, by transposing $\sin(\theta)$ to the other side and applying the identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$ to get the equation $\sin(\theta)(2\cos(\theta) – 1) = 0$ and then taking the general solution for $\sin(\theta) = 0$ and $2\cos(\theta) – 1 = 0$.
But I want to know how to use complex numbers to solve this question and get the same answers. I tried using Euler's formula and writing both $\sin(\theta)$ and $\sin(2\theta)$ as $(e^{i\theta}-e^{-i\theta})/2i$ and $(e^{2i\theta}-e^{-2i\theta})/2i$ respectively and I even got an equation of the form $y^4 – y^3 + y – 1 = 0$, where $y = e^{i\theta}$. However, solving that only yielded one of the principle solutions i.e $\theta = \pi$, I should also get $\theta = \pi/3$, but I haven't been able to get it.
Any help on this is greatly appreciated!
Best Answer
So, you tried to solve$$y+\frac1y=y^2+\frac1{y^2}.$$This is indeed equivalent to$$y^4-y^3+y-1=0.$$But\begin{align}y^4-y^3+y-1=0&\iff(y-1)(y^3+1)=0\\&\iff y=1(=e^0)\vee y=-1\left(=e^{\pi i}\right)\vee y=e^{i\pi/3}\vee y=e^{5\pi i/3}.\end{align}And so, yes, you do get the solutions $0$, $\pi$, $\frac\pi3$, and $\frac{5\pi}3$. Actually the whole set of solutions of the equation $\sin(\theta)=\sin(2\theta)$ is$$\pi\Bbb Z\cup\left\{\frac\pi3+2n\pi\,\middle|\,n\in\Bbb Z\right\}\cup\left\{\frac{5\pi}3+2n\pi\,\middle|\,n\in\Bbb Z\right\}.$$