Apply the quotient rule to find
$$ y' = \frac{e^x \cos x - e^x \sin x}{e^{2x}} $$
We want $y'=0$, which means we want the numerator to equal zero:
$$ 0 = e^x \cos x - e^x \sin x $$
$$ 0 = \cos x - \sin x $$
$$ \sin x = \cos x $$
Divide by $\cos x$:
$$ \tan x = 1 $$
This happens at $x = \frac{\pi}{4}$, and since tangent has period $\pi$, it will happen also when we add an integer multiple of $\pi$. Thus, we have a horizontal tangent line at
$$ x = \frac{\pi}{4} + n\pi, n \in \mathbb{Z} $$
The gradient$(m)$ of the tangent line $=f'(x)$
The tangent line will be horizontal of $y=f(x)$ if $f'(x)=0$ and will be vertical if $\displaystyle f'(x)=\infty\implies \frac1{f'(x)}=0$
Now, here $\displaystyle y=\sin2x+2\sin x\implies \frac{dy}{dx}=2\cos2x+2\cos x$
So, horizontal tangent, we need $\cos2x+\cos x=0\iff \cos2x=-\cos x=\cos(\pi-x)$
$\displaystyle\implies 2x=2n\pi\pm(\pi-x)$ where $n$ is any integer
Taking the '+' sign, $\displaystyle\implies 2x=2n\pi+(\pi-x)\implies x=\frac{(2n+1)\pi}3$
Now $0\le x<2\pi\implies 0\le \frac{(2n+1)\pi}3<2\pi\iff 0\le 2n+1<6\iff0\le n\le2$
Similarly, for the '-' sign
Now as $\displaystyle|\cos y|\le1$ for real $\displaystyle y, \frac{dy}{dx}=2\cos2x+2\cos x$ will be finite, hence no vertical tangent line
In fact, we can find the range of $\displaystyle 2\cos2x+2\cos x$ for real $x$
Best Answer
The derivative is cos(2x) we find the values of x when cos(2x) = 0 since the x is being multiplied by 2, this means there are twice as many answers for when cos(x) would equal 0. This means, using the unit circle, we can find the radians for which cos(x) = 0 and then take that radian and divide that by 2. Then just do an additional rotation, since it has twice as many answers, to get all four x values. Alternatively, you can use a graph to determine the answers by looking when y touches 0
NOTE: Since the interval is in the negative, this means rotations are also negative. In other words, all answers should be negative. Although you can make every answer you get negative and get away with it, you must be careful if the interval isn't a full rotation (i.e -π) because (-π/2) is different from (π/2) despite their similar values