Trying to use the sum-to-product formula to solve $\sin(2\theta)+\sin(4\theta)=0$ over the interval $[0,2\pi)$, but I'm missing solutions.
$$\sin(2\theta)+\sin(4\theta)=0$$
Apply sum-to-product formula:
$$2\sin\left(\frac{2\theta+4\theta}{2}\right)\cos\left(\frac{2\theta-4\theta}{2}\right)=0$$
$$2\sin(3\theta)\cos(-\theta)=0$$
By odd-even identities: $\cos(-\theta)=\cos(\theta)$
$$2\sin(3\theta)\cos(\theta)=0$$
$$\sin(3\theta)\cos(\theta)=0$$
By the zero-product property
$\sin(3\theta)=0$ or $\cos(\theta)=0$
Then solving for theta gives: $\theta=0, \frac{\pi}{2}, \frac{3\pi}{2}, \pi$.
However, there are missing solutions $\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$.
A solution online used double angle identities instead:
$$\sin(2\theta)+\sin(4\theta)=0$$
$$\sin(2\theta)+\sin(2*2\theta)=0$$
Apply double angle identity for: $\sin(2*2\theta)$
$$\sin(2\theta)+2\sin(2\theta)\cos(2\theta)=0$$
Factor out $\sin(2\theta)$
$$\sin(2\theta)*[1+2\cos(2\theta)]=0$$
Apply double angle identities:
$\cos(2\theta)= 1-2\sin^2(\theta)$
$\sin(2\theta)= 2\sin(\theta)\cos(\theta)$
$$2\sin(\theta)\cos(\theta)*[1+2(1-2\sin^2(\theta))]=0$$
$$2\sin(\theta)\cos(\theta)*[-4\sin^2(\theta)+3]=0$$
By the zero-product property
$2\sin(\theta)\cos(\theta)=0$ or $-4\sin^2(\theta)+3=0$
Which further simplifies to
$\sin(\theta)=0$, $\cos(\theta)=0$, or $-4\sin^2(\theta)+3=0$
Solving for theta now gives all possible solutions over $[0, 2\pi)$.
My questions are:
(1) Can the sum-to-product formula be used to solve this equation?
(2) If so, why were solutions missing when using the sum-to-product formula but not the double angle identities? What was I doing incorrectly?
Best Answer
This is an excellent way to proceed with this problem, and the reduction to $\sin(3\theta)\cos(\theta)=0$ is great; this implies that $\sin(3\theta)=0$ or $\cos(\theta)=0$.