Question

A set of values of x, satisfying the equation ${\cos }^2\left(\frac{1}{2}px\right)+{\cos }^2\left(\frac{1}{2}qx\right)=1$ form an Arithmetic progression with common difference

• A.
• B.
• C.
• D. none

Answer 1

The correct option is D none

1 + cos px + 1 + cos qx = 2
$\Rightarrow \cos \left(\frac{p+q}{2}\right)x\cos \left(\frac{p-q}{2}\right)x=0$
$\Rightarrow x=\frac{\left(2n+1\right)\pi }{p+q}$ or $\frac{\left(2n+1\right)\pi }{p-q}$
for n = 0, $\pm 1,\pm 2,.........$
forms an AP with common difference $\frac{2\pi }{p+q}$ or $\frac{2\pi }{p-q}$