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

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Answer 1 · 2015-08-22T22:28:48

Express $cos 2x$ in terms of $cos^2 x$ , then solve the resulting quadratic to find $cos x = 1$ , hence $x = 2n pi$ for any integer $n$ .

$cos 2x = cos^2 x - sin^2 x = cos^2 x - (1 - cos^2 x)$

$= 2 cos^2 x - 1$

So our equation becomes:

$2 cos^2 x - 1 = 4 cos x - 3$

Subtract the right hand side from the left to get:

$0 = 2 cos^2 x - 4 cos x + 2$

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

$=2(cos x - 1)(cos x - 1)$

So $cos x = 1$

So $x = 2n pi$ where $n$ is any integer.

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