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Q.
If $\cos x \cos y-\cos (x+y)=\frac{3}{2}$, then
Trigonometric Functions
Solution:
$\cos x+\cos y-\cos (x+y)=\frac{3}{2}$
or $2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)-2 \cos ^{2}\left(\frac{x+y}{2}\right)+1=\frac{3}{2}$
or $2 \cos ^{2}\left(\frac{x+y}{2}\right)-2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)+\frac{1}{2}=0$
Now, $\cos \left(\frac{x+y}{2}\right)$ is always real, then discriminant $\geq0$ .
Thus, $4 \cos ^{2}\left(\frac{ x - y }{2}\right)-4 \geq 0$
or $\cos ^{2}\left(\frac{ x - y }{2}\right) \geq 1$
or $\cos ^{2}\left(\frac{ x - y }{2}\right)=1$
or $\frac{x-y}{2}=0$ or $x=y$