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Answer 1 · 2017-12-06T17:53:59

$0, \frac{2 π}{3}, \frac{4 π}{3}$ $0,(2pi)/3, (4pi)/3$

Explanation:

Using identity:

$cos (2 x) = 2 {cos}^{2} x - 1$ $cos(2x)=2cos^2x-1$

$2 {cos}^{2} x - 1 - cos x = 0$ $2cos^2x-1-cosx=0$

Let $cos x = m$ $cosx=m$

$2 m^{2} - 1 - m = 0$ $2m^2-1-m=0$

Factor:

$(2 m + 1) (m - 1) = 0 \Rightarrow m = 1 and m = - \frac{1}{2}$ $(2m+1)(m-1)=0=>m=1 and m=-1/2$

Substitute $m = cos x$ $m=cosx$

$cos x = 1$ $cosx=1$

$cos x = - \frac{1}{2}$ $cosx=-1/2$

$cos x = 1 \Rightarrow x = 0$ $cosx=1=>x=0$

$cos x = - \frac{1}{2} \Rightarrow x = \frac{2 π}{3}, \frac{4 π}{3}$ $cosx=-1/2=>x=(2pi)/3, (4pi)/3$

Answer 2 · 2017-12-09T20:05:35

$0; \frac{2 π}{3}; \frac{4 π}{3}; π$ $0; (2pi)/3; (4pi)/3; pi$

Explanation:

cos 2x - cos x = 0
Replace cos 2s by $(2 {cos}^{2} x - 1)$ $(2cos^2 x - 1)$
$2 {cos}^{2} x - cos x - 1 = 0 .$ $2cos^2 x - cos x - 1 = 0.$
Solve this quadratic equation for cos x.
Since a + b + c = 0, use shortcut. The 2 real roots are:
cos x =1 and $cos x = \frac{c}{a} = - \frac{1}{2}$ $cos x = c/a = - 1/2$
a. cos x = 1
Unit circle gives 2 solutions:
x = 0, and $x = 2 π$ $x = 2pi$
b. $cos x = - \frac{1}{2}$ $cos x = -1/2$
Trig table and unit circle give 2 solutions
$x = \pm \frac{2 π}{3}$ $x = +- (2pi)/3$
The arc $- \frac{2 π}{3}$ $- (2pi)/3$ is co-terminal to arc $\frac{4 π}{3}$ $(4pi)/3$
Answers for $[0, 2 π]$ $[0, 2pi]$ :
$0; \frac{2 π}{3}; \frac{4 π}{3}; 2 π$ $0; (2pi)/3; (4pi)/3; 2pi$

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