Question

Solve the following equations.
$sin2x(0.1-(0.1\left)cosx\right)=sin2x+0.2si{n}^{3}x.$

Answer 1

$sin2x(0.1-(0.1\left)cosx\right)=sin2x+0.2si{n}^{3}x$

$\Rightarrow \left(0.1\right)-\left(0.1\right)cosx=\frac{sin2x+0.2si{n}^{3}x}{sin2x}$

we know that

$sin2x=2sinxcosx$

$(0.1-0.1cosx)=\frac{2sinxcosx+0.2si{n}^{3}x}{2sinxcosx}$

$0.1-0.1cosx=1+0.1\frac{si{n}^{2}x}{cosx}$

$0.1-0.1cosx=\frac{cosx+0.1si{n}^{2}x}{cosx}$

using cross multiplication

$(0.1-0.1cosx)cosx=cosx+0.1si{n}^{2}x$

$0.1\times cosx-0.1co{s}^{2}x=cosx+0.1si{n}^{2}x$

$0.1\times cosx=cosx+0.1(si{n}^{2}x+co{s}^{2}x)$

$1\{\because si{n}^{2}x+co{s}^{2}x=1\}$

$0.1\times cosx=cosx+0.1\times 1$

$0.1cosx-cosx=0.1$

$cosx(0.1-9)=0.1$

$cosx(-8.9)=0.1$

$cosx=-\frac{1}{89}$

$x=co{s}^{-1}(-\frac{1}{89})$

$x=\pi -co{s}^{-1}\left(\frac{1}{89}\right)$