Question

I don’t know how to solve this

Answer 1

Answer:

$\theta =2\pi k,\ \ k\in Z\ \\\text{or}\ \\\theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z$

Step-by-step explanation:

Given:

$\cos \theta-\sqrt{3}\sin \theta=1$

Divide this equation by 2:

$\dfrac{1}{2}\cos \theta-\dfrac{\sqrt{3}}{2}\sin \theta=\dfrac{1}{2}$

Note that

$\cos \dfrac{\pi }{3}=\dfrac{1}{2}\\ \\\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$

So, the previous equation is

$\cos \dfrac{\pi}{3}\cdot \cos \theta-\sin \dfrac{\pi}{3}\cdot \sin \theta=\dfrac{1}{2}$

Remind that

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

then

$\cos \left(\dfrac{\pi}{3}+\theta\right)=\dfrac{1}{2}$

The solution of this equation is

$\dfrac{\pi}{3}+\theta=\pm \arccos \dfrac{1}{2}+2\pi k,\ \ k\in Z\\ \\\dfrac{\pi}{3}+\theta=\pm \dfrac{\pi}{3}+2\pi k,\ \ k\in Z\\ \\\theta =2\pi k,\ \ k\in Z\ \text{or}\ \theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z$