Question

Solve the following equations.
$(1+cos4x)sin2x=co{s}^{2}2x.$

Answer 1

$(1+cos4x)sin2x=co{s}^{2}2x$

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

$\Rightarrow 2co{s}^{2}2x.sin2x-co{s}^{2}2x=0$

$\Rightarrow co{s}^{2}2x[2sin2x-1]=0$

$\therefore co{s}^{2}2x=0$ or $2sin2x-1=0$

$\Rightarrow 2x=(2x+1)\frac{\pi }{2}\Rightarrow sin2x=1/2$

$2x=(-1{)}^m\pi /6+m\pi$

$\Rightarrow x(2x+1)\pi /4\Rightarrow x=\frac{m\pi }{2}+(-1{)}^m\frac{\pi }{12}$

$nϵ$ integer

$mϵ$ integer