need help forgot how to do trig question

[hcky16]

I tried doing this problem but I don't think it is right can someone help me?
2(sin(x))^2+3sin(x)=-1 over the interval [0,2pi)

[eumyang]

Show us what you've tried, and then we might be able to help you.

[hcky16]

I tried,

2(sin(x))^2+3sin(x)=-1

2(sin(x))+3sin(x)+1=0

x=(-3(+/-)sqr 9-4(2)(1))/2(2)

x=(-3(+/-)1)/4

sinx=(-3(+/-)1)/4

x=arcsin(-3(+/-)1)/4

x=-90 or -30

x=-1.571 or -.524

[eumyang]

I tried,

2(sin(x))^2+3sin(x)=-1

2(sin(x))+3sin(x)+1=0

x=(-3(+/-)sqr 9-4(2)(1))/2(2)

x=(-3(+/-)1)/4
There's a typo in the 2nd line, and the last two lines aren't technically correct. You need to put the "sin" in front of the x. Also, you really didn't need to use the quadratic formula. This expression on the LHS:
2\sin^2 x + 3\sin x + 1 = 0
is factorable.

sinx=(-3(+/-)1)/4

x=arcsin(-3(+/-)1)/4

x=-90 or -30

x=-1.571 or -.524
Some problems here. First, these answers are not in the interval [0, 2pi). Just add 2pi to these answers and you'll be okay.

Second, the range of the arcsin function is only [-pi/2, pi/2], so there are actually 3 solutions that I see, not 2. (There is another angle whose sin is -1/2.)

[Sourabh N]

The question asks you to find the solution in [0,2pi). The answer you report is not.
For what value of x in [0,2pi) is sin(x) = -1 OR -1/2?