How do you solve #Cos (2theta) + 6sin^2 theta = 4#?

Question

How do you solve #Cos (2theta) + 6sin^2 theta = 4#?

1 Answer

Impact of this question

10409 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-26T04:34:05

#pi/3, (2pi)/3, (4pi)/3, (5pi)/3#

Explanation:

Use trig identity:
#cos 2t = 1 - 2sin^2 t.#
The equation is transformed to:
#1 - 2sin ^2t + 6sin^2 t = 4#
#4sin^2 t = 3#
#sin^2 t = 3/4#
#sin t = +- sqrt3/2#

Trig table and trig unit circle -->
a. #sin t = sqrt3/2# --> #t = pi/3# and #t = (2pi)/3#
b. #sin t = - sqrt3/2# --> #t = (4pi)/3# and #t = (5pi)/3#

Science

Math

Humanities

... and beyond

How do you solve #Cos (2theta) + 6sin^2 theta = 4#?

1 Answer

Explanation:

Related questions

Impact of this question