Question

The set of solutions of the system of equations :
$x+y=\frac{2\pi }{3}$ and
$\cos x+\cos y=\frac{3}{2}$
where x and y are real is

• A. $\left\{\right(x,y):\cos \left(\frac{x-y}{2}\right)=\frac{1}{2}\}$
• B. $\left\{\right(x,y):\sin \left(\frac{x-y}{2}\right)=\frac{1}{2}\}$
• C. $\left\{\right(x,y):\cos (x-y)=\frac{1}{2}\}$
• D. empty set

Answer 1

The correct option is D empty set

We have $x+y=\frac{2\pi }{3}$ and

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

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

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

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

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

Now maximum value of LHS is $=\sqrt{1^2+(\sqrt{3}{)}^2}=2$ Note: $-\sqrt{a^2+b^2}\le a\sin x+b\cos x\le \sqrt{a^2+b^2}$

Hence there is no solutions