Solve each equation, giving a general formula for all of the solutions:

Question

a. 2sin^2(x)-5 sin(x)-3=0
b. (cos(x)-((sqrt2)/2))(sec(x)-1)=0

Answer 1

just factor the first one, then there will be two values between 0 and 2π, and all multiples of 2π added thereto.
2sin^2(x)-5 sin(x)-3=0
(2sinx+1)(sinx-3) = 0
sinx = -1/2
sinx = 3 (not a valid solution)

The other is already factored:
(cos(x)-((√2)/2))(sec(x)-1)=0
(cosx-1/√2)(secx-1) = 0
cosx = 1/√2
secx = 1