Question

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

Answer 1

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

Explanation:

`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`