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!