Given $\cos x + 3\sin x = \sqrt{10} \cos(x-71.6)$, find the second solution in the interval $0 < \theta < 90$ $$\cos x + 3\sin x = \sqrt{10} \cos(x-71.6)$$
I've proven the above true. I must now solve this:
$$\cos 2\theta + 3\sin 2\theta = 2, 0<\theta<90$$
Here's what I've done so far:
$$\alpha = 2\theta - 71.6$$
$$\cos\alpha = \frac{2}{\sqrt{10}}$$
$$\alpha = 50.77$$
Now I've found $\theta = 61.2$, but there's another solution -- $\theta = 10.4$.
I don't understand how to find the second solution. The cos graph does not become positive again until the $270-360$ range, so the fourth quadrant, and they've restricted me to $0 < \theta < 90$.
Can someone explain (in some detail if possible) why and how to find my second solution? Does that same process work in general for problems like this? I often miss roots for the less simple trig angle business.

$\cos 2x + 3\sin 2x = \sqrt 10\cos (2x - \arctan 3)=2\\
\cos (2x - \arctan 3) = \frac {2}{\sqrt 10}\\
2x - \arctan 3 = \arccos\frac {2}{\sqrt 10}\\
x = \frac 12 (\arccos \frac {2}{\sqrt {10}} + \arctan 3)$
Regarding the second solution.
Since $\cos x = \cos -x$
Then $\alpha = -\arccos \frac {2}{\sqrt 10}$ will point to the second solution.
$x = \frac 12 (-\arccos \frac {2}{\sqrt {10}} + \arctan 3)$
Is also a solution.
