$\sin(4\theta) = \sin(2\theta+\pi/2)$ solve for $\theta$ when $\theta \in [0,\pi/4]$

Question

According to Wolfram, doing the inverse on both sides gives $4\theta = 2\theta + \pi/2 + 2\pi n_1$, makes sense, but it also gives $4\theta = -2\theta + \pi/2+2\pi n_2$. This is what i don't understand, why do you also get a minus?

Answer 1

If $\sin x = \sin y$, then either $x=y+2\pi n_1$, or $x=\pi-y+2\pi n_2$.

Now put $x=4\theta$ and $y=2\theta+\pi/2$.

Answer 2

Use that $$\sin(x)-\sin(y)=2 \sin \left(\frac{x}{2}-\frac{y}{2}\right) \cos \left(\frac{x}{2}+\frac{y}{2}\right)$$

Answer 3

Hint: your equation is equivalent to $\cos 2\theta (2\sin 2\theta -1)=0$. Note that $\arcsin\frac12=\frac{\pi}{6}$.