A simple trigonometric equation - need help solving it! ?

Question

I'm really not sure on how to solve this one? 

Find all the values of x in the interval [0, pi] which satisfy the equation cos(2x) = sin^2(x) 

So far i've tried doing this: 

cos(2x) = sin^2(x)
cos(2x) = 1 - cos^2(x)
cos^2(x) + cos(2x) - 1 = 0 

Now that we have a quadratic equation we apply the discriminant formula 

D = b^2 - 4ac
D = 1 - 4(1)(-1)
D = 1+ 4
D = 5 

so 

x1 = (1 + sqrt(5)) / (2)
x2 = (1  - sqrt(5)) / (2)

Is the process correct so far? And what do i do next?

Astral Walker · Accepted Answer

Your error is in the fact that you did not have a quadratic equation (you can't use cos(2x) in place of cos(x),

Instead use a basic identity for cos(2x)

cos(2x) = cos^2(x) - sin^2(x) = 1 - 2sin^2(2x)

So we have

1 - 2sin^2(x) = sin^2(x)

3sin^2(x) = 1

sin^2(x) = 1/3

sin(x) = +/-sqrt(1/3)

Now find all points in [0,pi] that satisfies this.  

Hint: Since sin(x) is bijective over [0,pi] you should find 1 solution for sin(x) = sqrt(1/3) and 1 solution for sin(x) = -sqrt(1/3)

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A simple trigonometric equation - need help solving it! ?

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