# Solving $\sin(ka/2)\cos(ka/2) = -\sin(ka)\cos(ka)$.

This is a basic high-school trigonometry problem, but it's been so long since I dealt with any of the trigonometric identities I am completely lost on how to solve it (I really should keep up with my high-school maths, this is embarassing!).

How do we go about solving (for $k$): $$\sin\left(\frac{ka}{2}\right)\cos\left(\frac{ka}{2}\right)=-\sin(ka)\cos(ka)$$

I recognise that we can reduce this to:

$$\sin(ka) = -\sin(2ka)$$

But I'm unsure how to proceed from here!

The equation may be written as $$\sin\left(ka\right)+2\sin\left(ka\right)\cos\left(ka\right)=0$$ or, equivalently, $$\sin\left(ka\right)\left(1+2\cos\left(ka\right)\right)=0$$ Can you proceed?
Using $$\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$$ we have $$\sin(ka)+\sin(2ka)=0\implies 2\sin\frac{3ka}{2}\cos\frac{ka}{2}=0$$ and so $$\sin\frac{3ka}{2}=0\quad\text{or}\quad \cos\frac{ka}{2}=0$$ I think that you can continue from here.