Solving $2\sin(\theta + 17) = \frac {\cos (\theta +8)}{\cos (\theta + 17)}$

For $0<\theta<360$

$$2\sin(\theta + 17) = \dfrac {\cos (\theta +8)}{\cos (\theta + 17)}$$

$$\Longrightarrow \sin(2\theta + 34)= \sin (82-\theta)$$

since sine is an odd function

$$2\theta + 34 = 82-\theta + 360n$$

This gives $\theta = 16, 136, 256$

But in the solution paper I can also see $64$. Why is it missing from the above method?

I can get $64$ (due to $\cos$)when I use $$\sin(\alpha) - \sin(\beta) = 2\sin\left(\dfrac {\alpha - \beta} 2\right)\cos\left(\dfrac {\alpha + \beta} 2\right)$$

But the first method I use should still work no?

The general solution to the equation:

$$\sin(x)=\sin(\alpha)$$

is actually given by the following:

$$x=n\pi+(-1)^n\alpha~\forall~n\in\Bbb Z$$

It is true that $\sin(2\theta + 34)= \sin (82-\theta)$ if $2\theta + 34 = 82-\theta + 360n$.
It is not true that $\sin(2\theta + 34)= \sin (82-\theta)$ only if $2\theta + 34 = 82-\theta + 360n$.
Remember that $\sin(180^\circ-\alpha) = \sin\alpha$. Thus if $\sin(2\theta+34^\circ) = \sin(82^\circ-\theta+n360^\circ)$ then $$\sin(2\theta+34^\circ) = \sin(180^\circ-\Big(82^\circ-\theta+n360^\circ\Big))$$ and you'll get additional solutions.
You also need to use the fact that $\sin x = \sin (180 - x)$.