Doubt regarding trigonometric equation

Question

In a book of mine it says solution of $\sin^2(x) = \sin^2(y)$ is $x = n\pi \pm y$

But if we take sq root on both sides we get sinx = siny for which the solution is

$x = n\pi + (-1)^ny$

Which is correct?

Answer 1

The book is right because you get $\sin(x)=\pm\sin(y)$ instead of $\sin(x)=\sin(y)$.

For example, the angles $\vartheta$ with $\sin^2(\vartheta)=\sin^2(10^{\circ})$ are $10^{\circ}$, $-10^{\circ}$, $10^{\circ}+180^{\circ}$ and $-10^{\circ}+180^{\circ}$, check it on a sketch!

Answer 2

Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,

$$\sin^2x-\sin^2y=\sin(x+y)\sin(x-y)$$ and $\displaystyle\sin A=0\iff A=n\pi$ where $n$ is any integer

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Alternatively, $$\cos2B=1-2\sin^2B$$

$$\implies\cos2x=\cos2y\iff2x=2m\pi\pm2y$$ where $m$ is any integer