How do you solve #2sin^2 x + sinx=1# for x in the interval [0,2pi)?
1 Answer
Dec 3, 2017
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]#
