Question

Solve the following equations.
$(cos6x-1)cot3x=sin3x.$

Answer 1

$(cos6x-1)cot3x=sin3x$

$\Rightarrow (cos6x-1)\frac{cos3x}{sin3x}=sin3x$

$\Rightarrow (cos6x-1)cos3x=si{n}^{2}3x$

$\Rightarrow (2co{s}^{2}3x-2)cos3x=1-co{s}^{2}3x$

Substitute

$cos3x=t$

$\Rightarrow 2(t^2-1)t=1-t^2$

$\Rightarrow 2t^3-2t=1-t^2$

$\Rightarrow 2t^3+t^2-2t-1=0$

$t=1$ is solution of the above equation

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

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

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

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

$\therefore t=1$ or $t=-1$ or $t=-1/2$

$\Rightarrow cos3x=1$ or $cos3x=-1$ or $cos3x=-1/2$

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

$x=\frac{2n\pi }{3}\Rightarrow x=(2k+1)\frac{\pi }{3}\Rightarrow x=\frac{2m\pi }{3}\pm \frac{2{\pi }^3}{9}$

$xϵz$ $kϵz$ $mϵz$