Question

Solve 0≤θ<π
2sin2θ=3(cos2θ+sinθ−1)

Answer 1

Let \( \sin(x) = u \) then \( \cos (x) = \sqrt{1-u^2}\)
\[ 2 \sin 2\theta = 3 ( \cos 2\theta + \sin \theta - 1) \\
4 u \sqrt{1 - u^2} = 3(1 - 2u^2 + u -1)\]
solve this equal for \( u \)

Answer 2

I managed to get x=\[\sin^{-1} \left( \sqrt{\frac{ 1 }{ 6 }} \right)\]
Which then gave me x=24 is this correct?

Answer 3

x=0, is for sure from here
http://www.wolframalpha.com/input/?i=Solve+4+x+sqrt%281-x%5E2%29+%3D+3+%28+1+-+2x%5E2+%2B+x-1%29
the other one is beyond the given range

Answer 4

Thanks, I just hope this doesn't come up in the exam as it was quite time consuming