Question #dc75d

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Answer 1 · 2017-02-10T05:58:10

$x=(7pi)/6 and (11pi)/6" when "x in [0,2pi]$

$2sin^2x-9sinx=5$
$=>2sin^2x-9sinx-5=0$

$=>2sin^2x-10sinx+sinx-5=0$

$=>2sinx(sinx-5)+1(sinx-5)=0$

$=>(sinx-5)(2sinx+1)=0$

$sinx=5->"not possible"$

$2sinx+1=0$

$=>sinx=-1/2=-sin(pi/6)$

$=>sinx=sin(pi+pi/6)=sin((7pi)/6)$
So $x=(7pi)/6$

Again

$=>sinx=-1/2=-sin(pi/6)$

$=>sinx=sin(2pi-pi/6)=sin((11pi)/6)$

$=>x=(11pi)/6$

Hence solutions are

$x=(7pi)/6 and (11pi)/6" when "x in [0,2pi]$