Confusion in Trigonometry Problems [closed]

Why can't I square both side and take $1$ to other side to make $2\cos^2x-1$ which is equal to $\cos2x$ and other side i would be left with $\sin2x$ and then I have $\tan2x=1$. Why this is wrong procedure I am not able to find answer this way. On the internet, it shows $\tan x = \pm\sqrt2-1$

Question:

The Principal Solution Set of the equation $$2\cos x = \sqrt{2+2\sin2x}$$ is

(i) $\{\pi/8,13\pi/8\}$

(ii) $\{\pi/4,13\pi/8\}$

(iii) $\{\pi/4,13\pi/10\}$

(iv) $\{\pi/8,13\pi/10\}$

This is a multiple choice question, and such questions frequently have quick solutions by ruling out impossible solutions.

Because of the positive square root we know that $\cos x\ge0$, which means $x$ cannot lie on the interval $\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$ which eliminates $\dfrac{13\pi}{10}$ as a solution.

So the correct choice must be either option (i) or (ii).

Since $\frac{\pi}{4}$ is not a solution, that leaves (i) as the correct choice.

ADDENDUM: This is not a nice exercise to work directly on a multiple choice type test, but we can divide both sides by $\sqrt{2}\cos x$ to get

\begin{eqnarray} \sqrt{2}&=&\sqrt{\sec^2x+2\tan x}\\ &=&\sqrt{\tan^2x+2\tan x+1}\\ &=&|\tan x+1| \end{eqnarray}

This gives $\tan x=-1\pm\sqrt{2}$ which does not translate to commonly known special angles, at least in introductory classes.

If $2\cos(x) = \sqrt{2+2\sin(2x)}$, then both sides must be positive.

Now squaring both sides gives

$$4\cos^2(x) = 2 + 2\sin(2x)$$

and applying the double angle identity to the left side,

$$2 + 2\cos(2x) = 2 + 2\sin(2x)$$

Solve for $x$:

$$\begin{align} 2 + 2\cos(2x) &= 2 + 2\sin(2x) \\ 2\cos(2x) &= 2\sin(2x) \\ \tan(2x) &= 1 \\ 2x &= \tan^{-1}(1) + n\pi \\ x &= \frac12 \tan^{-1}(1) + \frac{n\pi}2 \\ x &= \frac\pi8 + \frac{n\pi}2 \end{align}$$

where $n$ is any integer. This means the solutions to the equation belong to the set $$\left\{\ldots,-\frac{11\pi}8,-\frac{7\pi}8,-\frac{3\pi}8,\frac\pi8,\frac{5\pi}8,\frac{9\pi}8,\ldots\right\}$$ but some of these solutions are redundant; modulo $2\pi$, the solution set reduces to $$\left\{\frac\pi8,\frac{5\pi}8,\frac{9\pi}8,\frac{13\pi}8\right\}$$

Since $2\cos(x)$ must be positive, we can only allow angles $x$ that terminate in either the first or fourth quadrants; only $x=\frac\pi8$ and $x=\frac{13\pi}8$ meet this condition.