How do you solve cos2theta + 6sin^2theta = 2cos2θ+6sin2θ=2?

2 Answers
Mir Tareq
Jun 13, 2015

θ is 30 deg or 120 deg

Explanation:

cos2θ + 6sin²θ = 2 or, cos²θ − sin²θ + 6sin²θ = 2 or, cos²θ + sin²θ + 4sin²θ = 2 or, 1 + 4sin²θ = 2 or, sin²θ = 1/4 or, sinθ = +-1/2 or, θ = 30 deg, 120 deg

Nghi N
Mar 2, 2018

pi/6, (5pi)/6;(7pi)/6; (11pi)/6

Explanation:

cos 2t + 6sin^2 t = 2
Replace cos 2t by (1 - 2sin^2 t) --> trig identity:
1 - 2sin^2 t + 6sin^2 t = 2
4sin^2 t = 1
sin^2 t = 1/4
sin t = +- 1/2
Trig table and unit circle give 4 solutions:
a. sin t = 1/2 -->
t = pi/6 + 2kpi, and
t = (5pi)/6 + 2kpi
b. sin t = - 1/2 -->
t = - pi/6, or, as co-terminal, t = (11pi)/6 + 2kpi
t = pi - (-pi/6) = (7pi)/6 + 2kpi

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