Question
Solve sin^2x - 2sinx - 3 = 0 on the interval xE [0, 2pie]
Answers
Steve
pi, not pie
sin^2x - 2sinx - 3 = 0
(sinx-3)(sinx+1) = 0
sinx=3 nope
sinx = -1
x = 3pi/2
sin^2x - 2sinx - 3 = 0
(sinx-3)(sinx+1) = 0
sinx=3 nope
sinx = -1
x = 3pi/2
Sara
4cos2x - cosx-3 =0
0 to 2pi
0 to 2pi