When is $\sin(x) = \cos(x)$?

Here's an alternative hint with a more geometrical flavour: what are the angles in an isosceles right-angled triangle?

Hint: Take the equation $$ \sin(x) = \cos(x) $$ and divide both sides by $\cos(x)$ to get $$ \tan(x) = 1 $$

Alternatively, using a sum-to-product formula, we can observe that $$ \sin(x) - \cos(x) = \sqrt{2}\sin(x - 45^\circ) $$

A direct approach: use the unit-circle definition of sine and cosine.

Consider a unit circle around the origin of a Cartesian plane. Since in this problem $x$ is already in use as an angle, we cannot label the two axes $x$ and $y$ as usual, so let's label them $u$ (on the horizontal axis) and $v$ (on the vertical axis) instead. Then the unit-circle definition says that if we take a point at an angle $x$ radians counterclockwise around the unit circle, the coordinates at that point will be \begin{align} u &= \cos x, \\ v &= \sin x. \end{align}

The equation $\cos x = \sin x$ then tells us that $u = v$, which is the equation of a line at $\frac\pi4$ radians ($45$ degrees) through the origin. But we got these coordinates in the first place as coordinates of a point on the unit circle, so the solution must be at a point where the line $u = v$ intersects the unit circle. Draw a graph, as in the figure below: there are two such points.

The coordinates of these points happen to be $\left(\frac{\sqrt2}{2},\frac{\sqrt2}{2}\right)$ and $\left(-\frac{\sqrt2}{2},-\frac{\sqrt2}{2}\right)$, but even without figuring that out, since you know the angle of the line $u=v$ you can easily see the two possible values of $x$ within the range $0$ to $2\pi$ radians ($0$ to $360$ degrees).

1) $\cos^2 x + \sin^2 x = 1$

So $2 \cos^2 x = 1$

So $\cos x = \sin x = \pm \sqrt{\frac 12}$

2) $\sin x$ is the adjacent side of a right triangle. $\cos x $ is the opposite side. $\sin x = \cos x$ means the triangle is isoceles. So the base angles are congruent. So $x + x + 90 = 180$.

3) $\sin x = \cos (\frac {\pi}2 -x)$

so $\cos x = \sin x = \cos (\frac {\pi}2-x)$

Can you figure it out now?

There are some quadrant issues to figure out but they aren't hard.

If you have an equation of the form $$ \cos f(x)=\sin g(x) $$ you can rewrite it as $$ \cos f(x)=\cos\left(\frac{\pi}{2}-g(x)\right) $$ and so $$ f(x)=\frac{\pi}{2}-g(x)+2k\pi \qquad\text{or}\qquad f(x)=-\frac{\pi}{2}+g(x)+2k\pi $$

In your particular case, $f(x)=g(x)=x$, so you have $$ x=\frac{\pi}{2}-x+2k\pi \qquad\text{or}\qquad x=-\frac{\pi}{2}+x+2k\pi $$ Of course, the second possibility gives no solution; the first case gives $$ x=\frac{\pi}{4}+k\pi $$

If your function is $$ f(x)=\frac{\tan x}{(1+\tan x)^2} $$ the derivative is \begin{align} f'(x) &=\frac{(1+\tan^2x)(1+\tan x)^2-2(1+\tan x)(1+\tan^2x)\tan x} {(1+\tan x)^4} \\[6px] &=\frac{(1+\tan^2x)(1-\tan x)}{(1+\tan x)^3} \end{align} so it vanishes for $\tan x=1$.