Question

How do you solve the equation `2(sin(t))^2−sin(t)−1=0` in the interval [0, 2 ]? Give the answer in terms of pi. Thank You!!!!?

Answer 1

I would set `sin(t)=u` and write:
`2u^2-u-1=0`
`u_(1,2)=(1+-sqrt(1+8))/4=(1+-3)/4`
So, going back to `sin`
`sin(t)=1` for `t=pi/2`
`sin(t)=-1/2` for `t=7/6pi` but this I think it is outside your interval from 0 to 2 radians (unless you meant `2pi`, in which case it is ok).