Question

Tan(x+pi) + 2sin(x+pi)=0

can 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=(\sin x)/(\cos x)`

and so

`\tan x-2\sin x=\sin x\left(\frac1{\cos x}-2\right)`

which is only 0 if either
`\sin x=0` (which only happens for certain values of
`x`) or
`\frac1{\cos x}-2=0\implies\cos x=\frac12` (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<2\pi`, then
`\cos x=\frac12` is true for
`x=\frac\pi3` and
`x=\frac{5\pi}3`. Then to account for all other possible values, we add a multiple of
`2\pi`, so that
`x=\frac\pi3+2n\pi` or
`x=\frac{5\pi}3+2n\pi` for integers
`n`.