Question

Solve the following equations.
$4sin2x-3sin(2x-\frac{\pi }{2})=5.$

Answer 1

$\Rightarrow 4sin2x-3sin(2x-\frac{\pi }{2})=5$

$\Rightarrow 4sin2x-3sin[-(\frac{\pi }{2}-2x)]=5$

$\Rightarrow 4sin2x+3sin(\frac{\pi }{2}-2x)=5\{\because sin(-\theta )=-sin\theta \}$

$\Rightarrow 4sin2x+3cos2x=5....\{\because sin(\frac{\pi }{2}-\theta )=cos\theta \}$

$\Rightarrow 4(2sinx.cosx)+3(1-2si{n}^{2}x)=\mathrm{5...}\{\because sin2\theta =2sin\theta .cos\theta \because cos2\theta =1-2sin2\theta \}$

$\Rightarrow 8sinx.cosx+3-6si{n}^{2}x=5$

$\Rightarrow 8sinx.cosx-6si{n}^{2}x-2=0$

Divide both sides by $co{s}^{2}x$ we get

$\Rightarrow \frac{8sinx.cosx}{co{s}^{2}x}-\frac{6si{n}^{2}x}{co{s}^{2}x}-\frac{2}{co{s}^{2}x}=0$

$\Rightarrow 8tanx-6ta{n}^{2}x-2se{c}^{2}x=0$

$\Rightarrow 8tanx-6ta{n}^{2}x-2(1+ta{n}^{2}x)=\mathrm{0...}\{\because se{c}^2\theta =1+ta{n}^2\theta \}$

$\Rightarrow 8tanx-6ta{n}^{2}x-2-2ta{n}^{2}x=0$

$\Rightarrow 8tanx-8ta{n}^{2}x-2=0$

$\Rightarrow 8ta{n}^{2}x-8tanx+2=0$

$\Rightarrow 8ta{n}^{2}x-4tanx-4tanx+2=0$

$\Rightarrow 4tanx(2tanx-1)-2(2tanx-1)=0$

$\Rightarrow (2tanx-1)(4tanx-2)=0$

$\Rightarrow 2tanx-1=0$ or $4tanx-2=0$

$\therefore tanx=1/2$ or $\therefore tanx=1/2$

$\Rightarrow tanx=1/2$

$\Rightarrow x=ta{n}^{-1}\left(\frac{1}{2}\right)$