How do you find the general solutions for cos2x = 4cosx - 3cos2x=4cosx−3?

1 Answer
George C.
Aug 22, 2015

Express cos 2xcos2x in terms of cos^2 xcos2x, then solve the resulting quadratic to find cos x = 1cosx=1, hence x = 2n pix=2nπ for any integer nn.

Explanation:

cos 2x = cos^2 x - sin^2 x = cos^2 x - (1 - cos^2 x)cos2x=cos2x−sin2x=cos2x−(1−cos2x)

= 2 cos^2 x - 1=2cos2x−1

So our equation becomes:

2 cos^2 x - 1 = 4 cos x - 32cos2x−1=4cosx−3

Subtract the right hand side from the left to get:

0 = 2 cos^2 x - 4 cos x + 20=2cos2x−4cosx+2

=2(cos^2 x - 2 cos x + 1)=2(cos2x−2cosx+1)

=2(cos x - 1)(cos x - 1)=2(cosx−1)(cosx−1)

So cos x = 1cosx=1

So x = 2n pix=2nπ where nn is any integer.

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