Question

Solve the following equations.
$\sqrt{6-sinx-7co{s}^{2}x}+sinx=0.$

Answer 1

$\sqrt{6-sinx-7co{s}^{2}x}+sinx=0$

$\Rightarrow \sqrt{6-sinx-7co{s}^{2}x}+=-sinx$

Squaring both sides and subsitituting $t=sinx$

$\Rightarrow 6-t-7(1-t^2)=t^2$

$\Rightarrow 6-t-7+7t^2=t^2$

$\Rightarrow 6t^2-t-1=0$

$\Rightarrow 6t^2-3t+2t-1=0$

$\Rightarrow 3t(2t-1)+1(2t-1)=0$

$\Rightarrow (3t+1)(2t-1)=0$

$\therefore t=-1/3$ or $t=1/2$

$\Rightarrow sinx=-1/3$ $sinx=1/2$

$x=x\pi +(-1{)}^{n}si{n}^{-1}(-1/3)$ $x=m\pi +(-1{)}^m\pi /6$

$nϵz$ $mϵz$