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

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.

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

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

• Have you tried using the cos(2x) identity to get everything in the form of sin(x) May 1, 2019 at 20:15
• @rhombicosicodecahedron I'm not sure what you mean. May 1, 2019 at 20:16
• $Cos(2x) = 1-2sin^2(x)$ solution will further involve substitution May 1, 2019 at 20:17
• @rhombicosicodecahedron I have tried that just now. It gets me $2-4\sin^2(x)-\sin(x)=0$, but I'm not sure how to deal with the $2$. May 1, 2019 at 20:20
• try $\frac{pi}{2}$ for 2cos(x)+sin(2x) May 1, 2019 at 20:20

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 $$y=\sin x$$).
Hint:...... $$\cos2x=1-2\sin^2x$$
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. $$x=\frac{\pi}{6}$$.
Rewrite the equation as $$\;\cos(2x)=\cos \bigl(\frac\pi 2-x\bigr)$$.
Then use that $$\cos\alpha=\cos\beta\iff \alpha\equiv\pm\beta \pmod{2\pi}.$$