Find the least positive value of alpha for the following trignometric equation Given question is : $$\sin(5\alpha + \theta) = \cos(\theta - 3\alpha)$$
We are to find the least positive value of $\alpha$ for which above equation holds.
The way I did is as, $$\sin5\alpha \cos\theta + \cos5\alpha \sin\theta = \cos\theta \cos3\alpha + \sin\theta \sin3\alpha$$
Now for this to be true $$\sin5\alpha = \cos3\alpha$$ and $$\cos5\alpha = \sin3\alpha$$
How do I find the value of $\alpha$ that satisfies the above criteria?

My suggestion would be:
$$\sin(5\alpha + \theta) = \cos(\theta - 3\alpha)=\sin(\pi/2-\theta+3\alpha)$$
so, 
$$\sin(5\alpha + \theta) -\sin(\pi/2-\theta+3\alpha)=0$$
$$2\sin\left(\frac{(5\alpha + \theta)-(\pi/2-\theta+3\alpha)}{2}\right)\cdot \cos\left(\frac{(5\alpha + \theta)+(\pi/2-\theta+3\alpha)}{2}\right)=0$$
$$2\sin\left(\alpha + \theta-\frac{\pi}{4}\right)\cdot \cos\left(4\alpha+\frac{\pi}{4}\right)=0$$
So your full solution is:
$$\cos\left(4\alpha+\frac{\pi}{4}\right)=0\to 4\alpha+\frac{\pi}{4}=\frac{\pi}{2}+k\pi\to \alpha=\frac{\pi}{16}+\frac{k\pi}{4}$$
or
$$\sin\left(\alpha + \theta-\frac{\pi}{4}\right)=0\to \alpha + \theta-\frac{\pi}{4}=k\pi\to \alpha=\frac{\pi}{4}-\theta+k\pi$$
