posted by Maggie
Hello! Can someone please check and see if I did this right? Thanks! :)
Directions: Find the exact solutions of the equation in the interval [0,2pi]
that is not an answer. It is a restatement of the problem, using only sines.
2sin^2x-sinx-1 == 0
(2sinx+1)(sinx-1) = 0
sinx = -1/2 or 1, so
x = π/2, 4π/3, 5π/3