Working with LHS:
I've tried using the sum to product trig ID to get:
$1 + 2\cos(3x/2)\cos(x/2)$ from here I've tried a couple of things, but can't seem to get closer. I've tried changing the $(3x/2)$ into $(5x/2 – x)$ and using sum identity, but this just makes things even messier.
I also tried working the RHS. I'm only allowed to use the basic trig ID's: pythag, double and half angle, and sum to product and product to sum.
Best Answer
For $\sin\frac{x}{2}\neq0$ we obtain: $$1+\cos{x}+\cos2x=\frac{2\sin\frac{x}{2}+2\sin\frac{x}{2}\cos{x}+2\sin\frac{x}{2}\cos2x}{2\sin\frac{x}{2}}=$$ $$=\frac{2\sin\frac{x}{2}+\sin\frac{3x}{2}-\sin\frac{x}{2}+\sin\frac{5x}{2}-\sin\frac{3x}{2}}{2\sin\frac{x}{2}}=\frac{1}{2}+\frac{\sin\frac{5x}{2}}{2\sin\frac{x}{2}}.$$ I used the following formula. $$\sin\alpha\cos\beta=\frac{1}{2}(\sin(\alpha+\beta)+\sin(\alpha-\beta)).$$ For example,$$2\sin\frac{x}{2}\cos{x}=2\cdot\frac{1}{2}\left(\sin\left(\frac{x}{2}+x\right)+\sin\left(\frac{x}{2}-x\right)\right)=\sin\frac{3x}{2}-\sin\frac{x}{2}.$$