# Thread: [SOLVED] Correct double angle working?

1. ## [SOLVED] Correct double angle working?

The question asks to solve the following for $\displaystyle x$

$\displaystyle cos(2x)cos(\frac{\pi}{6})-sin(2x)sin(\frac{\pi}{6})=\frac{1}{2}$ where $\displaystyle x\in\left [0,2\pi\right ]$

My working:

$\displaystyle cos(2x)cos(\frac{\pi}{6})-sin(2x)sin(\frac{\pi}{6})=cos(2x+\frac{\pi}{6})$

$\displaystyle \Rightarrow cos(2x+\frac{\pi}{6})=\frac{1}{2},\ x\in\left [0,2\pi\right ]$

$\displaystyle 2x+\frac{\pi}{6}=\frac{\pi}{3},\ 2\pi-\frac{\pi}{3},\ 2\pi+\frac{\pi}{3},\ 4\pi-\frac{\pi}{3}$

$\displaystyle x=\frac{\pi}{12}$, $\displaystyle x=\frac{3\pi}{4}$, $\displaystyle x=\frac{7\pi}{4}$ and $\displaystyle x=\frac{13\pi}{12}$

I've been told it's wrong, with no explanation as to why.... As usual, I'm probably missing something really obvious?

* Edit - nvm, just had it rechecked and apparently my working is correct...

2. $\displaystyle \cos (2x+\frac{\pi}{6} ) = \frac{1}{2}$

you know that
$\displaystyle \cos x = \frac{1}{2}$

$\displaystyle x = \frac{\pi}{3} + 2k\pi$ ...(1)

$\displaystyle x= \frac{-\pi}{3} + 2k\pi$ ...(2)

so

$\displaystyle 2x + \frac{\pi}{6} = \frac{\pi}{3} + 2k\pi$

$\displaystyle x = \frac{\pi}{12} + k\pi$

but $\displaystyle x \in \left[0, \pi\right]$

so $\displaystyle x = \frac{\pi}{12} , \frac{\pi}{12} + \pi$
this form the first one

leave the rest for you