Question

$\text{The general value of x satisfying the equation}\sqrt{3}sinx+cosx=\sqrt{3}isgivenby$

• A. $x=n\pi +(-1{)}^n\frac{\pi }{4}+\frac{\pi }{3},n\in Z$
• B. $x=n\pi +(-1{)}^n\frac{\pi }{3}+\frac{\pi }{6},n\in Z$
• C. $x=n\pi \pm \frac{\pi }{6},n\in Z$
• D. $x=n\pi \pm \frac{\pi }{3},n\in Z$

Answer 1

The correct option is B

$x=n\pi +(-1{)}^n\frac{\pi }{3}+\frac{\pi }{6},n\in Z$

$Given:\sqrt{3}sinx+cosx=\sqrt{3}...\left(i\right)\text{This equation is of the form a}sin\theta +bcos\theta =c,wherea=\sqrt{3}b=1andc=\sqrt{3}Let:a=rcos\alpha andb=rsin\alpha Now,r=\sqrt{a^2+b^2}=\sqrt{(\sqrt{3}{)}^2}+1^2=2andtan\alpha =\frac{b}{a}\Rightarrow \tan \alpha =\frac{1}{\sqrt{3}}\Rightarrow \alpha =\frac{\pi }{6}Onputing\alpha =\sqrt{3}=rcos\alpha andb=1=rsin\alpha inequation\left(i\right),weget:rcos\alpha sinx+rsin\alpha cosx=\sqrt{3}\Rightarrow rsin(x+\alpha )=\sqrt{3}\Rightarrow 2sin(x+\alpha )=\sqrt{3}\Rightarrow sin(x+\alpha )=\frac{\sqrt{3}}{2}\Rightarrow sin(x+\alpha )=sin\frac{\pi }{3}\Rightarrow sin(x+\frac{\pi }{6})=sin\frac{\pi }{3}\Rightarrow x=n\pi +(-1{)}^n\frac{\pi }{3}-\frac{\pi }{6},n\in Z$