How do you solve $sinx=-sin^2x$ in the interval $0<=x<=2pi$ ?

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Answer 1 · 2016-10-29T16:13:18

$x={0,pi,(3pi)/2}$ in the interval $0 <= x <= 2pi$

$sinx=-sin^2x$

$hArrsinx+sin^2x=0$

or $sinx(1+sinx)=0$

Hence as we are seeking solution in the interval $0 <= x <= 2pi$

either $sinx=0$ i.e. $x=0$ or $x=pi$

or $six_1=0$ i.e $sinx=-1$ i.e. $x=(3pi)/2$

How do you solve sinx=-sin^2xsinx=-sin^2x in the interval 0<=x<=2pi0<=x<=2pi?