Question

Solve the following equations.
$si{n}^{4}x+co{s}^{4}x-2sin2x+\frac{3}{4}si{n}^{2}2x=0.$

Answer 1

$si{n}^{4}x+co{s}^{4}x-2sin2x+\frac{3}{4}si{n}^{2}2x=0$

$\Rightarrow si{n}^{4}x+co{s}^{4}x+2si{n}^{2}x.co{s}^{2}x-2si{n}^{2}x-2sin2x+\frac{3}{2}si{n}^{2}2x=0$

$\Rightarrow (si{n}^{2}x+co{s}^{2}x{)}^2-\frac{1}{2}(2sinx.cosx{)}^2-2sin2x+\frac{3}{4}si{n}^{2}2x=0$

$\Rightarrow 1-\frac{1}{2}si{n}^{2}2x-2sin2x+\frac{3}{4}si{n}^{2}2x=0$

$\Rightarrow \frac{si{n}^{2}2x}{4}-2sin2x+1=0$

$\Rightarrow si{n}^{2}2x-8sin2x+4=0$

say $t=sin2x$

$\Rightarrow t^2-8t+4=0$

$t=\frac{8\pm \sqrt{64-16}}{2}$

$=\frac{6\pm \sqrt{48}}{2}$

$=\frac{8\pm 4\sqrt{3}}{2}$

$=4\pm 2\sqrt{3}$

$\therefore sin2x=4+2\sqrt{3}$ or $sin2x=4-2\sqrt{3}=2(2-\sqrt{3})$

Again $sin2xϵ[-1,1],$ Hence the only solution possible is

$sin2x=4-2\sqrt{3}$

$2x=x\pi +(-1{)}^{n}si{n}^{-1}(4-2\sqrt{3})$

$x=\frac{x\pi }{2}+(-1{)}^n\frac{si{n}^{-1}(4-2\sqrt{3})}{2}$