Squaring both sides leads to a complete different result

Question

I came across with this equation:

$v^2 \cos (2 \alpha) \sqrt{v^2 \sin^2 \alpha+2gh}+v \sin \alpha [v^2 \cos (2 \alpha)-2gh]=0$

or

$2 g h \sin \alpha - v^2 \cos(2\alpha) \sin \alpha = v \cos(2 \alpha) \sqrt{v^2 \sin^2 \alpha + 2gh}$

Now this equation before squaring gives $$\alpha = \arccos \left(\sqrt{\frac{2 g h + v^2}{2 g h + 2v^2}} \right)$$

But if we square its sides:

$4 g^2 h^2 \sin^2 \alpha-4ghv^2\sin^2 \alpha \cos(2\alpha)+v^4 \cos^2(2\alpha) \sin^2 \alpha = v^2 [\cos^2(2\alpha)](v^2 \sin^2 \alpha + 2gh)$

The solution would be:

$$\alpha=\arccos \left(\frac{gh}{2v^2+2gh} \right)$$

Which is a whole different result. I'm wondering why squaring ruins the result? I do know that this operation may introduce extra solutions, but here we have a complete new one.

Answer 1

We have $$v^2\cos2\alpha(\sqrt{v^2 \sin^2 \alpha+2gh}+v\sin\alpha)=2gh v\sin\alpha$$

Rationalizing the numerator & cancelling $v\ne0$

$$v\cos2\alpha\frac{2gh}{\sqrt{v^2 \sin^2 \alpha+2gh}-v\sin\alpha}=2gh \sin\alpha$$

Cancelling $2gh\ne0$ & using $\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

$$v(\cos^2\alpha-\sin^2\alpha)=\sin\alpha(\sqrt{v^2 \sin^2 \alpha+2gh}-v\sin\alpha)$$

$$\implies v\cos^2\alpha=\sin\alpha\sqrt{v^2 \sin^2 \alpha+2gh}$$

Squaring we get $$ v^2(1-\sin^2\alpha)^2=\sin^2\alpha(v^2 \sin^2 \alpha+2gh)$$

Cancel $\displaystyle v^2\sin^4\alpha$ and use $\displaystyle \cos^2\alpha=1-\sin^2\alpha$