f(x) = 4cos^2 x - 4sin x - 1 = 0f(x)=4cos2x−4sinx−1=0
Replace cos^2 xcos2x by (1 - sin^2 x)(1−sin2x)
4(1 - sin^2 x) - 4sin x - 1 = 04(1−sin2x)−4sinx−1=0
4 - 4sin^2 x - 4sin x - 1 = 04−4sin2x−4sinx−1=0
Change side. Solve the quadratic equation by the improved quadratic formula in graphic form (Socratic, Google Search):
4sin^2 x + 4sin x - 3 = 04sin2x+4sinx−3=0
D = d^2 = b^2 - 4ac = 16 + 48 = 64D=d2=b2−4ac=16+48=64 --> d = +- 8d=±8
There are 2 real roots:
sin x = - b/(2a) +- d/(2a) = - 1/2 +- 1sinx=−b2a±d2a=−12±1
sin x = - 1/2 + 1 = 1/2sinx=−12+1=12 and sin x = - 3/2sinx=−32 (rejected as < - 1)
sin x = 1/2sinx=12 --> trig table and unit circle give -->
x = pi/6x=π6, and x = pi - pi/6 = (5pi)/6x=π−π6=5π6
