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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.

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

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

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    $\begingroup$ Have you tried using the cos(2x) identity to get everything in the form of sin(x) $\endgroup$ May 1, 2019 at 20:15
  • $\begingroup$ @rhombicosicodecahedron I'm not sure what you mean. $\endgroup$ May 1, 2019 at 20:16
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    $\begingroup$ $Cos(2x) = 1-2sin^2(x)$ solution will further involve substitution $\endgroup$ May 1, 2019 at 20:17
  • $\begingroup$ @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 $2$. $\endgroup$ May 1, 2019 at 20:20
  • $\begingroup$ try $\frac{pi}{2}$ for 2cos(x)+sin(2x) $\endgroup$
    – ricky
    May 1, 2019 at 20:20

4 Answers 4

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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 $y=\sin x$).

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

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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. $x=\frac{\pi}{6}$.

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Hint:

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

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

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