I am stuck on this trig equation.

Solve the Equation for one function of $\displaystyle \theta$ and find the angle.

$\displaystyle sin\theta + 3 cos\theta = 2 \sqrt{2}$

ok, step one is write in terms of one function. so looking at it I can't see any easy identity to replace. so I raise power on both sides.

$\displaystyle

\begin{aligned}

sin\theta + 3 cos\theta = 2 \sqrt{2}&=(sin\theta + 3 cos\theta)^2 = (2\sqrt{2})^2 \\

&=sin^2\theta + 6cos\theta sin\theta+9cos^2\theta = 8 \\

&=sin^2\theta + 6cos\theta sin\theta+9(1-sin^2\theta) = 8 \\

&=8sin^2\theta - 6cos\theta sin\theta = 1

\end{aligned}

$

if I raise power again it will not square the other cos. and I can't see any elementary identity to use on it. in the book it says the answer should be $\displaystyle \frac{\pi}{4}, 45^{\circ}$