Solve the equation $cos 2 x - cos x = 0$ $cos2x-cosx=0$ ?

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Answer 1 · 2017-04-04T12:51:53

$x = (2 n + 1) π$ $x=(2n+1)pi$ or $x = 2 n π \pm \frac{2 π}{3}$ $x=2npi+-(2pi)/3$ , where $n$ $n$ is an integer

Using identity $cos 2 x = 2 {cos}^{2} x - 1$ $cos2x=2cos^2x-1$ , $cos 2 x - cos x = 0$ $cos2x-cosx=0$ can be written as

$2 {cos}^{2} x - cos x - 1 = 0$ $2cos^2x-cosx-1=0$ and hence using quadratic formula

$cos x = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 \times 2 \times (- 1)}}{2 \times 2}$ $cosx=(-(-1)+-sqrt((-1)^2-4xx2xx(-1)))/(2xx2)$

$= \frac{1 \pm \sqrt{9}}{4}$ $=(1+-sqrt9)/4$

$= \frac{1 \pm 3}{4}$ $=(1+-3)/4$

$= 1$ $=1$ or $- \frac{1}{2}$ $-1/2$

If $cos x = 1$ $cosx=1$ , $x = (2 n + 1) π$ $x=(2n+1)pi$ , where $n$ $n$ is an integer

and if $cos x = - \frac{1}{2}$ $cosx=-1/2$ , $x = 2 n π \pm \frac{2 π}{3}$ $x=2npi+-(2pi)/3$ , where $n$ $n$ is an integer

Solve the equation cos2x-cosx=0cos2x−cosx=0cos2x-cosx=0?