Question

Find the general solution of $x$ ${\cos }^{2}2x+{\cos }^{2}3x=1$

• A. $(2k+1)\frac{\pi }{10},k\in I$
• B. $(\pi +1)\frac{\pi }{10};k\in I$
• C. $(2k-1)\frac{\pi }{10},k\in I$
• D. Both (A) and (C)

Answer 1

The correct option is C $(2k-1)\frac{\pi }{10},k\in I$

general solution of

${\cos }^{2}2x+{\cos }^{2}3x=1$

we know

$\cos 2x=2{\cos }^{2}x-1$

${\cos }^{2}x=\frac{\cos 2x+1}{2}$

So in the given question

$\left(\frac{\cos 4x+1}{2}\right)+\left(\frac{\cos 6x+1}{2}\right)=1$

$\frac{\cos 4x}{2}+\frac{1}{2}+\frac{\cos 6x}{2}+\frac{1}{2}=1$

$\frac{\cos 4x}{2}=\frac{\cos 6x}{2}$

$\cos 4x=\cos (\pi +6x)$

we know

$\cos x=cosy$

then $x=(2n\pi +y)$

$4x=(2k\pi \pm (\pi +6x\left)\right)$

take $+$ $-$

$4x=2k\pi +\pi +6x$

$-2x=2k\pi +\pi$

$x=-k\pi -\frac{\pi }{2}$

$4x=2k\pi -\pi -6x$

$10x=(2k-1)\pi$

$x=(2k-1)\frac{\pi }{10}$