Question

Solve the following equations.
$8co{s}^{6}x=3cos4x+cos2x+4.$

Answer 1

$8co{s}^{6}x=3cos4x+cos2x+4$

$8co{s}^{6}x=3(2co{s}^{2}x-1)+2co{s}^{2}x-1+4$

[As $cos2x=2xo{s}^{2}x-1]$

$8co{s}^{6}x=6co{s}^{2}2x-3+2co{s}^{2}x-1+4$

$8co{s}^{6}x=6(2co{s}^{2}x-1{)}^2+2co{s}^{2}x$

solving we get

$4co{s}^{6}x-12co{s}^{4}x+11co{s}^{2}x-3=0$

take $co{s}^{2}x=t$

then,

$4t^3-12t^2+11t-3=\mathrm{0...}\left(1\right)$

Now put random value of t like $-1,0,-1$

If that satisfy $e{q}^n$ then root

$t=1$

$4(1{)}^3-12(1{)}^2+11\left(1\right)-3$

$=0$

then $t=1$ is a root of $e{q}^n$(1)

$(t-1)(4t^2-8t+3)=0$

$(t-1)(t^2-2t+3/4)=0$

$(t-1)(t-1/2)(t-3/2)=0$

then

$co{s}^{2}x=1$ OR $co{s}^{2}x=\frac{1}{2}$ OR $co{s}^{2}x-\frac{3}{4}$

$x=co{s}^{-1}(+1)$

$x=co{s}^{-1}(\pm \frac{1}{\sqrt{2}})x=co{s}^{-1}\left(\sqrt{\frac{3}{2}}\right)$

$t=\frac{2\pm \sqrt{4-4.3/4}}{2}$

$t=\frac{2\pm 1}{2}$

$t=3/2$ OR $1/2$