Question

How do you solve `2sin^2 x + sinx=1` for x in the interval [0,2pi)?

Answer 1

`pi/6,(5pi)/6,(3pi)/2`

Explanation:

`"rearrange euating to zero"`

`rArr2sin^2x+sinx-1=0`

`"we now have a quadratic in sin"`

`rArr(2sinx-1)(sinx+1)=0`

`rArr2sinx-1=0" or "sinx+1=0`

`rArrsinx=1/2" or "sinx=-1`

`rArrx=pi/6" or "(5pi)/6" or "(3pi)/2to[0,2pi]`