Question

Solve the following equations.
$sin2x+tanx=2.$

Answer 1

$sin2x+tanx=2$

$2sinxcosx+\frac{sinx}{cosx}=2$

$2sinxco{s}^{2}x+sinx=2cosx$

$(2co{s}^2+1)=2\frac{cosx}{sinx}$

Squaring both sides

$(4co{s}^{4}x+1+4co{s}^{2}x)=\frac{4co{s}^{2}x}{si{n}^{2}x}$

$si{n}^{2}x(4co{s}^{4}x+1+4co{s}^{2}x)=4co{s}^{2}x$

$(1-co{s}^{2}x)(4co{s}^{4}x+1+4co{s}^{2}x)=4co{s}^{2}x$

solving we get

$4co{s}^{6}x+co{s}^{2}x-1=0$

$\frac{t=co{s}^{2}x}{4t^3+t-1=0}$

then $t=1/2$ satisfies this $e{q}^n$ hence

$(t-1/2)(2t^2+t+1)=0$

$\downarrow$

+ve quality

Hence $co{s}^{2}x=1/2$

$x=co{s}^{-1}(\pm \frac{1}{\sqrt{2}})$ Ans