If $cos(2x) + cosx = 0$ , how would you find the value of $x$ on $[0, 2pi)$ ?

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Answer 1 · 2016-11-02T03:03:25

$x = pi/3, (5pi)/3, pi$

Use the identity $cos2x = 1 -2sin^2x$ to start the solving process.

$cos2x + cosx = 0$

$1 - 2sin^2x + cosx = 0$

Now, use the identity $sin^2theta = 1 - cos^2theta$ to get rid of the sine in the equation.

$1 - 2(1- cos^2x) + cosx = 0$

$1 - 2 + 2cos^2x + cosx = 0$

$2cos^2x + cosx - 1 = 0$

Let $t = cosx$ .

$2t^2 + t - 1 = 0$

$2t^2 +2t - t - 1 = 0$

$2t(t + 1) - (t + 1) = 0$

$(2t - 1)(t + 1) = 0$

$t = 1/2 and -1$

$cosx = 1/2 and cosx= -1$

$x = pi/3, (5pi)/3, pi$

Hopefully this helps!

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