Question

I don’t know how to solve this

Answer 1

`\theta =2\pi k,\ \ k\in Z\ \\\text{or}\ \\\theta=-(2\pi)/(3)+2\pi k,\ \ k\in Z`

Explanation:

Given:

`\cos \theta-√(3)\sin \theta=1`

Divide this equation by 2:

`(1)/(2)\cos \theta-(√(3))/(2)\sin \theta=(1)/(2)`

Note that

`\cos (\pi )/(3)=(1)/(2)\\ \\\sin (\pi )/(3)=(√(3))/(2)`

So, the previous equation is

`\cos (\pi)/(3)\cdot \cos \theta-\sin (\pi)/(3)\cdot \sin \theta=(1)/(2)`

Remind that

`\cos x\cos y-\sin x\sin y=\cos (x+y),`

then

`\cos \left((\pi)/(3)+\theta\right)=(1)/(2)`

The solution of this equation is

`(\pi)/(3)+\theta=\pm \arccos (1)/(2)+2\pi k,\ \ k\in Z\\ \\(\pi)/(3)+\theta=\pm (\pi)/(3)+2\pi k,\ \ k\in Z\\ \\\theta =2\pi k,\ \ k\in Z\ \text{or}\ \theta=-(2\pi)/(3)+2\pi k,\ \ k\in Z`