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.


Apply sum-to-product formula:



By odd-even identities: $\cos(-\theta)=\cos(\theta)$



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:



Apply double angle identity for: $\sin(2*2\theta)$


Factor out $\sin(2\theta)$


Apply double angle identities:

$\cos(2\theta)= 1-2\sin^2(\theta)$

$\sin(2\theta)= 2\sin(\theta)\cos(\theta)$



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?

  • $\begingroup$ No solutions were missing! $\endgroup$ – Andrew Chin Jul 2 '20 at 23:05
  • $\begingroup$ Wouldn't it be easier to use $$\sin(2\theta)+\sin(4\theta)= \sin(2\theta)+2\sin(2\theta)\cos(2\theta)=\sin(2\theta)\cdot(2+\cos(2\theta)?$$ $\endgroup$ – Michael Hoppe Jul 3 '20 at 11:45

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 $\cos(\theta)=0$.

  • The solutions to $\cos(\theta)=0$ are $\theta = \dots,\frac\pi2,\frac{3\pi}2,\dots$.
  • The solutions to $\sin(\alpha)=0$ are $\alpha = \dots, 0, \pi, 2\pi, \dots$. But we have $\sin(3\theta)=0$, and so the solutions are $3\theta = \dots, 0, \pi, 2\pi, \dots$, which is the same as $\theta=\dots,0,\frac\pi3,\frac{2\pi}3,\pi,\frac{4\pi}3,\frac{5\pi}3,\dots$.
  • $\begingroup$ Ah yes, I see my mistake now. Thank you. $\endgroup$ – Slecker Jul 2 '20 at 23:18

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