How do you find the exact value of 3cos2theta+costheta+2=03cos2θ+cosθ+2=0 in the interval 0<=theta<2pi0≤θ<2π?

1 Answer
Nghi N
Jul 13, 2017

x = +- 120^@x=±120∘
x = +- 70^@53x=±70∘53

Explanation:

Replace cos 2t by (2cos^2 t - 1)(2cos2t−1) -->
3(2cos^2 t - 1) + cos t + 2 = 03(2cos2t−1)+cost+2=0
Solve this quadratic equation for cos t
6cos^2 t + cos t - 1 = 06cos2t+cost−1=0
D = d^2 = b^2 - 4ac = 1 + 24 = 25D=d2=b2−4ac=1+24=25 --> d = +- 5d=±5
There are 2 real roots:
cos x = -b/(2a) +- d/(2a) = - 1/12 +- 5/12 = (-1 +- 5)/12cosx=−b2a±d2a=−112±512=−1±512
cos x = - 6/12 = -1/2cosx=−612=−12
cos x = 4/12 = 1/3cosx=412=13
Use calculator and unit circle:
a. cos x = - 1/2cosx=−12 --> x = +- 120^@x=±120∘
b. cos x = 1/3cosx=13 --> x = +- 70^@53x=±70∘53

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