Question

Solve the following equations.
$2sin3x-\frac{1}{sinx}=2cos3x+\frac{1}{cosx}.$

Answer 1

$2sin3x-\frac{1}{sinx}=2cos3x+\frac{1}{cosx}$

$\Rightarrow (2sin3x.sinx-1)/sinx=(2cos3x.cosx+1)/cosx$

$\Rightarrow 2sin3x.sinx.cosx-cosx=2cos3x.cosx.sinx+sinx$

$\Rightarrow sin3x.sin2x-cosx=cos3x.sin2x+sinx$

$2sin3x-\frac{1}{sinx}=2cos3x+\frac{1}{cosx}$

$\Rightarrow 2sin3x-2cos3x=1/sinx+1/cosx$

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

$\Rightarrow 6(sinx+cosx)-8(si{n}^{3}x+co{s}^{3}x)=(sinx+cosx)/sinx.cosx$

$\Rightarrow 6(sinx+cosx)-8(sinx+cosx)(1-sinx.cosx)$

$=(sinx+cosx)/sinx.cosx$

$\Rightarrow (sinx+cosx)[6-8(1-sinx.cosx)-\frac{1}{sinx.cosx}]=0$

$\Rightarrow (sinx+cosx)(8sinx.cosx-2-\frac{1}{sinx.cosx})=0$

$\therefore sinx+cosx=0$ or $sinx.cosx=1/2$ or $sinx.cosx=-1/4$

For $sinx+cosx=0$ or $sinx.cosx=1/2$

$\Rightarrow tanx=-1\Rightarrow sin2x=1$

$x=k\pi -\pi /4\Rightarrow 2x=2x\pi +\pi /2$

$kϵ$ integers.

$x=n\pi +\pi /4$

$nϵ$ integers

For $sinx.cosx=-1/4$

$\Rightarrow sin2x=-1/2$

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

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