Question

find all exact solutions in [0,2pi) of the trigonometric equation 1-2cos^2(x/2) = sin^2(x)

Answer 1

your idea?

Answer 2

what do you mean?

Answer 3

\[1-2\cos ^{2}(\frac{ x }{ 2 }) = \sin ^{2}(x) \in [0,2\pi)\]

Answer 4

I think you have some strategy to approach the problem, right? give me your idea. I have to know where you stuck

Answer 5

To be honest, I'm completely lost. Have no clue how to start

Answer 6

how about double angle rule? nothing?

Answer 7

double angle rule = a little

Answer 8

\[\sin 2u = 2 \sin u \cos u\]

Answer 9

ok, \[cos \dfrac{x}{2}= \pm\sqrt{\dfrac{1+cos x}{2}}\]
so that \( cos^2(x/2) =\dfrac{1+cos x}{2}\) --> \(2cos^2(x/2)= 1+cosx\)