trigonometry - Solve: $cos(theta + 25^circ) + sin(theta +65^circ) = 1$

This afternoon I've been studying the pythagorean identities & compound angles. I've got a problem with a question working with 2 sets of compound angles:

Solve, in the interval $0^\circ \leq \theta \leq 360^\circ$, $$\cos(\theta + 25^\circ) + \sin(\theta +65^\circ) = 1$$

I've attempted expanding but reach a point with no common factors & see how to manipulate the trig ratios to move on; is there a solution without expanding?

$$\cos\theta\cos 25^\circ-\sin\theta \sin 25^\circ+\sin\theta\cos 65^\circ+\sin 65^\circ\cos\theta=1$$

$$\cos \theta\;\left(\cos 25^\circ+\sin 65^\circ\right) +\sin\theta\;\left(\cos 65^\circ-\sin 25^\circ\right)=1$$

Could you tell me if I've made a mistake or how I could continue;
thanks

coffee is wearing out

Answer

"Solve, in the interval 0≤θ≤360, cos(θ+25)+sin(θ+65)=1"

$\cos(\theta+25)+\cos(25-\theta)=1$

∵ $\sin(\theta+65)=\cos(90-(\theta+65))=\cos(25-\theta)$

$\cos(\theta)\cos(25)- \sin(\theta)\sin(25)+\cos(25)\cos(\theta)+\sin(\theta)\sin(25)=1$

$2\cos(\theta)\cos(25)=1$

$\cos(\theta)=\frac{1}{2\cos(25)}$

∴$\theta≈55.6°, 326.5°$