How do you solve $2-2sin^2(x/2)=2cos^2(x)$ ?

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Answer 1 · 2015-05-02T03:14:18

Use the trig identity: $2.sin^2 a = (1 - cos 2a)$

Replace: $2.sin^2 (x/2) = (1 - cos x)$

$f(x) = 2cos^2 x + 1 - cos x - 2 = 0$ . Call cos x = t --> quadratic equation.

$f(t) = 2t^2 - t - 1 = 0$
Since a + b + c = 0, one real root is t = 1 and the other is $t = c/a = -1/2$

Next, solve t = cos x = 0 -> $x = Pi/2 and x = 3Pi/2$
Then, solve t = cos x = $-1/2$ -> $x = 2Pi/3 and x = 4Pi/3$

How do you solve 2-2sin^2(x/2)=2cos^2(x)2-2sin^2(x/2)=2cos^2(x)?