Expressing sin(2θ) and cos(2θ) in terms of x

[TrigEatsMe]

For the question,
If x = 3cos(θ), 0<θ<π/2, express sin(2θ) in terms of x.

I confirmed that the following is correct:
sin(2θ)=(2x/9)√(9-x^2)

But for this next one...the x+1 is throwing me through a loop or something.

If x+1 = 3sin(θ), 0<θ<π/2, express cos(2θ) in terms of x.

cos(2θ) = fail

Any tips would be appreciated. Thanks!

 

[Prove It]

For the question,
If x = 3cos(θ), 0<θ<π/2, express sin(2θ) in terms of x.

I confirmed that the following is correct:
sin(2θ)=(2x/9)√(9-x^2)

But for this next one...the x+1 is throwing me through a loop or something.

If x+1 = 3sin(θ), 0<θ<π/2, express cos(2θ) in terms of x.

cos(2θ) = fail

Any tips would be appreciated. Thanks!

$\displaystyle \begin{align*} \cos{ \left( 2\theta \right) } &= 1 - 2\sin^2{ \left( \theta \right) } \\ &= 1 - 2 \left( \frac{x + 1}{3} \right) ^2 \\ &= 1 - 2 \left( \frac{x^2 + 2x + 1}{9} \right) \\ &= 1 - \frac{2x^2 + 4x + 2}{9} \\ &= \frac{9}{9} - \frac{2x^2 + 4x + 2}{9} \\ &= \frac{7 - 4x - 2x^2}{9} \end{align*}$