Question

$Ifcos\theta +\sqrt{3}sin\theta =2,then\theta =$

• A. $\pi /3$
• B. $2\pi /3$
• C. $4\pi /3$
• D. $5\pi /3$

Answer 1

The correct option is A

$\pi /3$

$Given:cos\theta +\sqrt{3}sin\theta =2...\left(i\right)\text{This equation is of the form a}cos\theta +bsin\theta =c,wherea=1,b=\sqrt{3}andc=2Let:a=rcos\alpha ,\sqrt{3}=rsin\alpha \Rightarrow r=\sqrt{a^2+b^2}=\sqrt{1+3}=\sqrt{4}=2And,tan\alpha =\frac{b}{a}\Rightarrow tan\alpha =\frac{\sqrt{3}}{1}\Rightarrow tan\alpha =\sqrt{3}\Rightarrow \alpha =\frac{\pi }{3}Onputtinga=1=rcos\alpha andb=\sqrt{3}=rsin\alpha inequation\left(i\right),wegetrcos\theta cos\alpha +rsin\theta sin\alpha =2\Rightarrow rcos(\theta -\alpha )=2\Rightarrow 2cos(\theta -\frac{\pi }{3})=2\Rightarrow cos(\theta -\frac{\pi }{3})=1\Rightarrow cos(\theta -\frac{\pi }{3})=cos\theta \Rightarrow \theta -\frac{\pi }{3}=2n\pi \pm 0Forn=0,=\frac{\pi }{3}\therefore \theta =\frac{\pi }{3}$