Question

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

• A. $x=(2n+1)\frac{\pi }{4},nϵI$
• B. $x=(4n+1)\frac{\pi }{4},nϵI$
• C. $x=n\pi \frac{\pi }{4},nϵI$
• D. $noneofthese$

Answer 1

Answer: (A)

$x=(2n+1)\frac{\pi }{4},nϵI$

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

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

$(\cos x+\cos 3x{)}^2-2\cos x\cos 3x={\sin }^{2}2x$

$\therefore 2\cos 2x\cos x-2\cos x\cos 3x={\sin }^{2}2x$

$\therefore 2\cos x(\cos 2x-\cos 3x)={\sin }^{2}2x$

$\therefore 2\cos x(\cos 2x-\cos 3x)=4{\sin }^{2}x{\cos }^{2}x$

$\therefore 2\cos x(2{\cos }^{2}x-1-(4{\cos }^{3}x-3\cos x)=4{\sin }^{2}x{\cos }^{2}x$

$\therefore 4{\cos }^{3}x-2\cos x-8{\cos }^{4}x+6{\cos }^{2}x=4{\sin }^{2}x{\cos }^{2}x$

$\therefore 4{\cos }^{2}x-2-8{\cos }^{3}x+6\cos x=4{\sin }^{2}x\cos x$

$\therefore 4{\cos }^{2}x-2-8{\cos }^{3}x+6\cos x=4(1-{\cos }^{2}x)\cos x$

$\therefore 4{\cos }^{2}x-2-8{\cos }^{3}x+6\cos x=4\cos x-4{\cos }^{3}x$

$\therefore -4{\cos }^{3}x+2\cos x-2+4{\cos }^{2}x=0$

$\therefore 2{\cos }^{3}x-2{\cos }^{2}x-\cos x+1=0$

$\therefore 2{\cos }^{2}x(\cos x-1)-1(\cos x-1)=0$

$\therefore (2{\cos }^{2}x-1)(\cos x-1)=0$

$\therefore \cos x=\frac{+1}{\sqrt{2}}$ OR $\cos x=1$

$\therefore x=(2x+1)\frac{\pi }{4}$ $n\in I$.