How do you solve 2sin^2(x)+3cos(x)=02sin2(x)+3cos(x)=0 on the interval [0,2pi]?

1 Answer
Mar 20, 2016

(2pi)/3, (4pi/3)2π3,(4π3)

Explanation:

Replace in the equation sin^2 xsin2x by (1 - cos^2 x) (1cos2x)-->
2(1 - cos^2 x) + 3cos x = 02(1cos2x)+3cosx=0
-2cos^2 x + 3cos x + 2 = 02cos2x+3cosx+2=0
Solve this equation for cos x.
D = d^2 = b^2 - 4ac = 9 + 16 = 25D=d2=b24ac=9+16=25 --> d = +- 5d=±5
cos x = -b/(2a) +- d/(2a) = 3/4 +- 5/-4 = 3/4 +- 5/4cosx=b2a±d2a=34±54=34±54
a. cos x = 8/4 = 2cosx=84=2 (rejected because > 1)
b. cos x = -2/4 = -1/2cosx=24=12
cos x = -1/2cosx=12 --> x = +- (2pi)/3x=±2π3
Answers for the interval (0, 2pi)(0,2π):
(2pi)/3, (4pi)/32π3,4π3 (co-terminal to (-2pi)/3)2π3)