Question

The solution set of $x+y=\frac{2\pi }{3}$ and $\cos x+\cos y=\frac{3}{2}$ where x and y are real is

• A. $\varphi$ (empty)
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $\varphi$ (empty)

Given $x+y=\frac{2\pi }{3}$ and $\cos x+\cos y=\frac{3}{2}$

$y=(\frac{2\pi }{3}-x)$

$\Rightarrow \cos x+\cos (\frac{2\pi }{3}-x)=\frac{3}{2}$

$\Rightarrow \cos x+(-\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x)=\frac{3}{2}$

$\Rightarrow \frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x=\frac{3}{2}$

$\Rightarrow \sin \frac{\pi }{6}\cos x+\cos \frac{\pi }{6}\sin x=\frac{3}{2}$

$\Rightarrow \sin (x+\frac{\pi }{6})=\frac{3}{2}=1.5$ $[\because \sin (A+B)=\sin A\cos B+\sin B\cos A]$

Which is never possible as $-1\le \sin \theta \le 1$ but here $\sin (x+\frac{\pi }{6})=1.5$

Hence, no solution exists