Question

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

Answer 1

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`