Question

Solve (sin(x)+1)(2(sin(x)^2)-3sin(x)-2) on the interval [0,2pi).

Answer 1

I'm assuming (sin(x)+1)(2(sin(x)^2)-3sin(x)-2)=0. This is a graph I have drawn up...

Answer 2

Since it's already factorised you can just consider each bracket separately so you have \[\sin x +1 = 0 \qquad 2\sin^2 x -3\sin x - 2 = 0\]

Answer 3

@bfsgd For sin(x) = -1, I got -pi/2, and for 2sin(x)^2-3sin(x) = 2 I got (3 + or - i(sqrt7))/4. I think I am wrong but I don't know what to do.

Answer 4

For the first one you're right but the answer you give isn't in the interval (0,2pi). So the answer to the first one is....?

Answer 5

Also got 3pi/2.

Answer 6

Yep that's the one. I would say for the second bracket you're best saying that there are no solutions in the real numbers.

Answer 7

Well that was the x value for a quadratic equation where I plugged in a=2, b=-3, and c=-2.

Answer 8

Yes and the discriminant is negative so you get something in the complex numbers.

Answer 9

But I know there are supposed to be three solutions so...what now?