Question

$\sqrt{3}\cos \theta +\sin \theta =1;-2\pi <\theta <2\pi$

Answer 1

We have,

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

Since, $\sin \theta =\sqrt{1-{\cos }^2\theta }$

Therefore,

$(1-\sqrt{3}\cos \theta {)}^2=1-{\cos }^2\theta$

$1+3{\cos }^2\theta -2\sqrt{3}\cos \theta =1-{\cos }^2\theta$

$4{\cos }^2\theta -2\sqrt{3}\cos \theta =0$

$4\cos \theta =2\sqrt{3}$

$\cos \theta =\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$

$\theta ={\cos }^{-1}\frac{\sqrt{3}}{2}=\frac{\pi }{6}$ or $2\pi -\frac{\pi }{6}=\frac{11\pi }{6}$

But

$\theta =\frac{11\pi }{6}$ is only correct by putting in the equation.

Hence, this is the answer.