Question

Use compound-angle formulae to solve cos(x-30)=2cos(x+30), for 0<x<360 ?

Answer 1

`x=30^\circ, 210^\circ`

Explanation:

Given that

`\cos(x-30^\circ)=2\cos(x+30^\circ)`

`\cosx\cos(30^\circ)+\sin x\sinx(30^\circ)=2(\cosx\cos(30^\circ)-\sin x\sinx(30^\circ))`

`\cos x\cdot \sqrt3/2+\sin x\cdot 1/2=2(\cos x\cdot \sqrt3/2-\sin x\cdot 1/2)`

`\sqrt3 \cos x+\sin x\=2\sqrt 3\cos x-2\sin x\`

`\sin x+2\sin x=2\sqrt3\cos x-\sqrt3\cos x`

`3\sin x=\sqrt3\cos x`

`\frac{\sin x}{\cos x}=\sqrt3/3`

`\tan x=1/\sqrt3`

`\tan x=\tan(\pi/6)`

`x=k\pi +\pi/6`

Where `k` is any integer i.e. `k=0, \pm1, \pm2, \pm3, \ldots`

Since, `0 < x< \360^\circ` hence setting `k=0, 1`, we get desired values as follows

`x=30^\circ, 210^\circ`