How do you solve #cos2theta + 6sin^2theta = 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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