Question

Solve the following equations.
$(cos2x-1)co{t}^{2}x=-3sinx.$

Answer 1

$(cos2x-1)co{t}^{2}x=-3sinx$

$\Rightarrow (cos2x-1)co{t}^{2}x=-3sinx$

$\Rightarrow -(1-cos2x)co{t}^{2}x=-8sinx$

$\Rightarrow -2si{n}^{2}x\frac{co{s}^{2}x}{si{n}^{2}x}=-3sinx$

$\Rightarrow +2co{s}^{2}x=\ne 3sinx$

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

$\Rightarrow 2(1-t^2)=3t$ (say $sinx=t$ )

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

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

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

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

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

$sinx=1/2$ or $sinx=-2$

But $sinxϵ[-1,1]$ hence $sinx\ne -2$ for any $xϵ$ real number

$\therefore$ the only solution possible is

$sinx=1/2$

$\Rightarrow x=x\pi +(-1{)}^n\pi /6;nϵz$