Question

The general solution of the equation; $\cos x.\cos 6x=-1,$ is:

• A. $x=(2n+1)\pi$
• B. $x=2n\pi$
• C. $x=(2n-1)\pi$
• D. none of these

Answer 1

The correct option is A $x=(2n+1)\pi$

$\cos x.\cos 6x=-1$
$\cos x\left(\cos 2\right(3x\left)\right)=-1=\cos x(2{\cos }^{2}3x-1)$
$=\cos x\left(2\right(4{\cos }^{3}x-3\cos x{)}^2-1)=-1$
$=\cos x\left(2\right(16{\cos }^{6}x+9{\cos }^{2}x-24{\cos }^{4}x-1\left)\right)=-1$
$=32{\cos }^{7}x-48{\cos }^{5}x+18{\cos }^{3}x-\cos x+1=0$
$=(\cos x+1)(32{\cos }^{6}x-32{\cos }^{5}x-16{\cos }^{4}x+16{\cos }^{3}x+2{\cos }^{2}x-2cox+1)=0$
The only real solution for this is ,
$\cos x=-1$
So, the principle solution for this is $\pi$
General solution=>$2n\pi +\pi$