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Answer 1 · 2017-02-10T05:58:10

$x = \frac{7 π}{6} and \frac{11 π}{6} when x \in [0, 2 π]$ $x=(7pi)/6 and (11pi)/6" when "x in [0,2pi]$

$2 {sin}^{2} x - 9 sin x = 5$ $2sin^2x-9sinx=5$
$\Rightarrow 2 {sin}^{2} x - 9 sin x - 5 = 0$ $=>2sin^2x-9sinx-5=0$

$\Rightarrow 2 {sin}^{2} x - 10 sin x + sin x - 5 = 0$ $=>2sin^2x-10sinx+sinx-5=0$

$\Rightarrow 2 sin x (sin x - 5) + 1 (sin x - 5) = 0$ $=>2sinx(sinx-5)+1(sinx-5)=0$

$\Rightarrow (sin x - 5) (2 sin x + 1) = 0$ $=>(sinx-5)(2sinx+1)=0$

$sin x = 5 \to not possible$ $sinx=5->"not possible"$

$2 sin x + 1 = 0$ $2sinx+1=0$

$\Rightarrow sin x = - \frac{1}{2} = - sin (\frac{π}{6})$ $=>sinx=-1/2=-sin(pi/6)$

$\Rightarrow sin x = sin (π + \frac{π}{6}) = sin (\frac{7 π}{6})$ $=>sinx=sin(pi+pi/6)=sin((7pi)/6)$
So $x = \frac{7 π}{6}$ $x=(7pi)/6$

Again

$\Rightarrow sin x = - \frac{1}{2} = - sin (\frac{π}{6})$ $=>sinx=-1/2=-sin(pi/6)$

$\Rightarrow sin x = sin (2 π - \frac{π}{6}) = sin (\frac{11 π}{6})$ $=>sinx=sin(2pi-pi/6)=sin((11pi)/6)$

$\Rightarrow x = \frac{11 π}{6}$ $=>x=(11pi)/6$

Hence solutions are

$x = \frac{7 π}{6} and \frac{11 π}{6} when x \in [0, 2 π]$ $x=(7pi)/6 and (11pi)/6" when "x in [0,2pi]$