Question 1036084: For questions 1-2 solve each equation on the interval [0,2pi). Show all work.
1. 4 sin^2 x + 2 sin x- 2 = 0
2. 2 sin(3x) = 1
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 4 sin^2 x + 2 sin x- 2 = 0
factor out a 2.
2(2sin^2 x+ sin x -1)=0
(2sinx-1)(sin x +1)=0
2 sin x -1=0, 2 sin x=1, sin x =(1/2), where it is at pi/6 and 5 pi/6.
sin x+1=0
sin x=-1
That is the case at 3pi/2.
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2sin (3x)=1
sin (3x)=(1/2)
The solutions above are divided by 3, so it occurs at 3x=pi/6, and pi/18 and 5pi/18 would be the solutions.
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