Question

The number of values of $x\in (0,\pi )$ satisfying the equation $\sin x+\sqrt{3}\cos x=\sqrt{2}$ is

• A. $3$
• B. $1$
• C. $2$
• D. $0$

Answer 1

The correct option is B $1$

Given the equation: $\sin x+\sqrt{3}\cos x=\sqrt{2}$
Divinding both sides with $\sqrt{1^2+(\sqrt{3}{)}^2}=\sqrt{4}=2,$ we get the equation as:
$\Rightarrow (\sin x\times \frac{1}{2}+\frac{\sqrt{3}}{2}\cos x)=\frac{\sqrt{2}}{2}$
$\Rightarrow \sin x\times \cos \frac{\pi }{3}+\cos x\times \sin \frac{\pi }{3}=\frac{1}{\sqrt{2}}$
$\Rightarrow \sin (x+\frac{\pi }{3})=\frac{1}{\sqrt{2}}=\sin \frac{\pi }{4}$
$\Rightarrow x+\frac{\pi }{3}=n\pi +(-1{)}^n\frac{\pi }{4}\forall n\in Z$
Now, substituting value(s) of $n=0,1,2,3,\cdots$ we get:
$a.n=0\Rightarrow x+\frac{\pi }{3}=\frac{\pi }{4}\Rightarrow x=-\frac{\pi }{12}$
$b.n=1\Rightarrow x+\frac{\pi }{3}=\pi -\frac{\pi }{4}\Rightarrow x=\frac{5\pi }{12}$
$c.n=2\Rightarrow x+\frac{\pi }{3}=2\pi +\frac{\pi }{4}>\pi$
Thus, for $x\in (0,\pi ),$ we have only one solution.