Question

Solve the following equations.
$cos9x-2cos6x=2.$

Answer 1

$cos9x-2cos6x=2...\left(1\right)$

Now we can write

$cos9x=4co{s}^{3}3x-3cos3x$

$cos6x=2co{s}^{2}3x-1$

consider $cos3x=t$

equation (1) can be written as

$\therefore 4t^3-4t^2-3t+2=2$

$\Rightarrow t(4t^2-4t-3)=0$

$\Rightarrow t(4t^2-6t+2t-3)=0$

$\Rightarrow t\left(2t\right(2t-3)+1(2t-3\left)\right)=0$

$\Rightarrow t(2t+1)(2t-3)=0$

$\therefore t=0$ or $t=\frac{-1}{2}$ or $t=\frac{3}{2}$

$\Rightarrow cos3x=0$ or $t=\frac{-1}{2}$

$3x=\frac{(2n+1)\pi }{2}cos3x=\frac{-1}{2}$

$x=\frac{(2n+1)}{\frac{\pi }{6}}3x=2n\pi \pm \frac{2\pi }{3}$

$x=\frac{2m\pi }{2}\pm \frac{2\pi }{9}$

$xϵ$ integers

$mϵ$ integer

$t=\frac{3}{2}$ is not possible