# Solve the following equation for 0 less than or equal to x less than or equal to \pi: 2\cos^{2} x...

## Question:

Solve the following equation for {eq}0 \leq x \leq \pi\textrm{: } \quad 2\cos^{2} x - \cos x = 0 {/eq}.

Select all that apply.

(a) {eq}\displaystyle\; 0 {/eq}

(b) {eq}\displaystyle\; \frac{\pi}{3} {/eq}

(c) {eq}\displaystyle\; \frac{\pi}{2} {/eq}

(d) {eq}\displaystyle\; \frac{2\pi}{3} {/eq}

## Trigonometric Equations:

Trigonometric substitution is a variable change that can be used in solving equations to find the value of the argument of the trigonometric functions found in the equation.

## Answer and Explanation:

{eq}\begin{align*} 2\cos^2\left(x\right)-\cos\left(x\right)&=0 \\ \cos\left(x\right)&=h \\ 2h^2-h&=0 \\ 2h\left(h-\frac{1}{2}\right)&=0 \\ \\ \\ 2h&=0 \quad h=0 \\ \cos\left(x\right)&=0 \quad x=\cos^{-1}\left(0\right)=\frac{\pi}{2} \\ \\ \\ h-\frac{1}{2}&=0 \quad h=\frac{1}{2} \\ \cos\left(x\right)&=\frac{1}{2} \quad x=\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3} \end{align*} {/eq}

**Answer: **

The option that apply are the option b and option c.

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from Calculus: Help and Review

Chapter 3 / Lesson 6