Question

$Avalueof\theta satisfyingcos\theta +\sqrt{3}sin\theta =2is$

• A. $\frac{5\pi }{3}$
• B. $\frac{4\pi }{3}$
• C. $\frac{2\pi }{3}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is D

$\frac{\pi }{3}$

$Givenequation:cos\theta +\sqrt{3}sin\theta =2...\left(i\right)\text{Thus,the equation is of the form}\alpha cos\theta +bsin\theta =c,where\alpha =1,b=\sqrt{3}andc=3.Let:a=rcos\alpha andb=rsin\alpha 1=rcos\alpha and\sqrt{3}=rsin\alpha \Rightarrow r=\sqrt{a^2+b^2}=\sqrt{(\sqrt{3}{)}^2+1^2}=2andtan\alpha =\frac{b}{a}\Rightarrow tan\alpha =\frac{\sqrt{3}}{1}\Rightarrow tan\alpha =tan\frac{\pi }{3}\Rightarrow \alpha =\frac{\pi }{3}Onputting\alpha =1=rcos\alpha andb=\sqrt{3}=rsin\alpha inequation\left(i\right),weget:rcos\alpha cos\theta +rsin\alpha sin\theta =2\Rightarrow rcos(\theta -\alpha )=2\Rightarrow rcos(\theta -\frac{\pi }{3})=2\Rightarrow 2cos(\theta -\frac{\pi }{3})=2\Rightarrow cos(\theta -\frac{\pi }{3})=1\Rightarrow cos(\theta -\frac{\pi }{3})=cos0\Rightarrow \theta -\frac{\pi }{3}=0\Rightarrow \theta =\frac{\pi }{3}$