## Calculus 10th Edition

Published by Brooks Cole

# Chapter 2 - Differentiation - 2.4 Exercises - Page 137: 83

#### Answer

Three points: $(\dfrac{\pi}{6}, \dfrac{3\sqrt{3}}{2})$, $(\dfrac{5\pi}{6}, -\dfrac{3\sqrt{3}}{2})$ and $(\dfrac{3\pi}{2}, 0)$

#### Work Step by Step

$f(x)=g(x)+h(x)\rightarrow g(x)=2\cos{x}$ ; $h(x)=\sin{2x}$ $g'(x)=-2\sin{x}$; $h'(x)=2\cos{2x}$ $f'(x)=g'(x)+h'(x)=2\cos{2x}-2\sin{x}$ $f'(x)=0\rightarrow 2\cos{2x}-2\sin{x}=0$ To solve the trigonometric equation you can graph it (quick and efficient but not always possible) or algebraically as follows: $2(1-2\sin^2{x})-2\sin{x}=0\rightarrow$ $2\sin^2{x}+\sin{x}-1=0\rightarrow \sin{x}=\dfrac{1}{2}$ or $\sin{x}=-1\rightarrow$ $x=\dfrac{\pi}{6}$, $x=\dfrac{5\pi}{6}$ or $\dfrac{3\pi}{2}.$ By plugging each value into the original function we get three points in the specified domain: $(\dfrac{\pi}{6}, \dfrac{3\sqrt{3}}{2})$, $(\dfrac{5\pi}{6}, -\dfrac{3\sqrt{3}}{2})$ and $(\dfrac{3\pi}{2}, 0)$

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