Question

Tan(x+pi) + 2sin(x+pi)=0can you help prove this and make the left side equal the right?

Answer 1

This is not an identity but an equation. You're supposed to find the values of $x$ for which this is true. It is *not* true for all values of $x$.

To see why:

$\tan x$ has period $\pi$, which means $\tan(x+\pi)=\tan x$.

The angle sum identity for $\sin x$ says that

$\sin(x+\pi)=\sin x\cos \pi+\cos x\sin\pi=-\sin x$

So

$\tan(x+\pi)+2\sin(x+\pi)=\tan x-2\sin x$

By definition of $\tan x$,

$\tan x=\dfrac{\sin x}{\cos x}$

and so

$\tan x-2\sin x=\sin x\left(\dfrac1{\cos x}-2\right)$

which is only 0 if either $\sin x=0$ (which only happens for certain values of $x$) or $\dfrac1{\cos x}-2=0\implies\cos x=\dfrac12$ (which also only happens for certain values of $x$). It's these values of $x$ you want to find.

$\sin x=0$ whenever $x$ is a multiple of $\pi$, i.e. $x=n\pi$ for any integer $n$.

If $0\le x$, then $\cos x=\dfrac12$ is true for $x=\dfrac\pi3$ and $x=\dfrac{5\pi}3$. Then to account for all other possible values, we add a multiple of $2\pi$, so that $x=\dfrac\pi3+2n\pi$ or $x=\dfrac{5\pi}3+2n\pi$ for integers $n$.