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Find all solutions to the equation in the interval [0,2pi). Enter the solutions in increasing order. Cos 2x = cos x

Find all solutions to the equation in the interval [0,2pi). Enter the solutions in-example-1
User Folyd
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1 Answer

7 votes

Explanation:


\cos(2x) = \cos(x)


\cos {}^(2) (x) - \sin {}^(2) (x) = \cos(x)


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


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


2 \cos {}^(2) (x) - \cos(x) - 1 = 0


2 \cos {}^(2) (x) - 2 \cos(x) + \cos(x) - 1 = 0


2 \cos(x) ( \cos(x) - 1) + 1( \cos(x) + 1)


(2 \cos(x) + 1)( \cos(x) - 1) = 0


2 \cos(x) + 1 = 0


2 \cos(x) = - 1


\cos(x) = - (1)/(2)


x = (2\pi)/(3) ,x = (4\pi)/(3)


\cos(x) - 1 = 0


\cos(x) = 1


x = 0

So our answer are


0, (2\pi)/(3) , (4\pi)/(3)

User Sinix
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