How to find the $x$ values of $2\cos(2x)-2\sin(x)=0$?

Question

I am not sure how to find the $x$ values of $2\cos(2x)-2\sin(x)=0$.

I am trying to find the absolute values of $f(x)=2\cos(x)+\sin(2x)$ in the range$[0, \frac{\pi}{2}]$.

I have differentiated $f(x)$ to produce $2\cos(2x)-2\sin(x)$, but I am unsure how to find the zeros of the function since I cannot think of a value of $x$ that would make both sine and cosine equal 0.

$$\frac{d}{dx} 2\cos(x)=[0 \cdot \cos(x)] + [2 \cdot -\sin(x)]=-2\sin(x)$$ $$\frac{d}{dx}\sin(2x)= \cos(2x) (2)=2\cos(2x)$$

$$\frac{d}{dx}2\cos(x)+\sin(2x)=2\cos(2x)-2\sin(x)$$

Answer 1

You don't need any calculus to do this: $$ 0=2\cos(2x)-2\sin(x) = 2(\cos^2x -\sin^2x -\sin x) = 2(1-\sin x - 2\sin^2x). $$ which is a quadratic equation (substitute $y=\sin x$).

Answer 2

Hint:...... $$\cos2x=1-2\sin^2x$$

Answer 3

Let $s:=\sin x$ so $\sin x-\cos 2x=2s^2+s-1$, which $=0$ for $s\in\{-1,\,\frac12\}$. For $x\in[0,\,\frac{\pi}{2}]$, only $s=\frac12$ is obtainable, viz. $x=\frac{\pi}{6}$.

Answer 4

Hint:

Rewrite the equation as $\;\cos(2x)=\cos \bigl(\frac\pi 2-x\bigr)$.

Then use that $$\cos\alpha=\cos\beta\iff \alpha\equiv\pm\beta \pmod{2\pi}.$$