Question

Solve the following equations.
$si{n}^{4}x+co{s}^{4}x=sin2x-\mathrm{0.5.}$

Answer 1

$si{n}^{4}x+co{s}^{4}x=sin2x-1/2$

$\Rightarrow (si{n}^{2}x{)}^2+(co{s}^{2}x{)}^2=sin2x=-1/2$

$\Rightarrow (si{n}^{2}x{)}^2+(co{s}^{2}x{)}^2+2si{n}^{2}x.co{s}^{2}x-2si{n}^{2}x.co{s}^{2}x=sin2x-1/2$

$\Rightarrow (si{n}^{2}x+co{s}^{2}x{)}^2-2si{n}^{2}x.co{s}^{2}x=sin2x-1/2$

$\Rightarrow 1-2si{n}^{2}x.co{s}^{2}x=sinx-1/2$

$\Rightarrow 1-\frac{1}{2}(2sinx.cosx{)}^2=sin2x-1/2$

$\Rightarrow 1-1/2(sin2x{)}^2=sin2x-1/2$

say $sin2x=t$

$\Rightarrow 1-\frac{t^2}{2}=t\frac{-1}{2}$

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

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

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

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

$\therefore t=1$ or $t=-3$

$\Rightarrow sin2x=1$ or $sin2x=-3$

$\Rightarrow 2x=(4x+1)\frac{\pi }{2}$ (no solution possible as $sin2xϵ[-1,1])$

$\Rightarrow x=(4x+1)\frac{\pi }{4}$