How do you solve $2 {cos}^{2} (x) = 3 cos (x)$ $2cos^2 (x) = 3 cos (x)$ ?

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Answer 1 · 2015-05-14T20:32:44

If $cos (x) = 0$ $cos(x) = 0$ then $2 {cos}^{2} (x) = 3 cos (x) = 0$ $2cos^2(x) = 3 cos(x) = 0$ , so that gives solutions of the form:

$x = \frac{π}{2} + n π$ $x = pi/2 + n pi$ , where $n$ $n$ is any integer.

On the other hand, if $cos (x) \neq 0$ $cos(x) != 0$ then divide both sides of the equation by $2 cos (x)$ $2cos(x)$ to get:

$cos x = \frac{3}{2}$ $cos x = 3/2$

Since $- 1 \leq cos x \leq 1$ $-1 <= cos x <= 1$ for all values of $x$ $x$ , this provides no more solutions.

How do you solve 2cos^2 (x) = 3 cos (x)2cos2(x)=3cos(x)2cos^2 (x) = 3 cos (x)?