How do you find all the solutions in the interval [0, 2π) without using a calculator of #3sinx = 2cos^2x#?

Question

How do you find all the solutions in the interval [0, 2π) without using a calculator of #3sinx = 2cos^2x#?

1 Answer

Impact of this question

10543 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-24T05:01:26

Solve 3sin x = 2cos^2 x

Ans: #pi/6 and (5pi)/6#

Explanation:

#3sin x = 2cos^2 x#
#3sin x = 2( 1 - sin^2 x) = 2 - 2sin^2 x#
#2sin^2 x + 3sin x - 2 = 0#
Solve this quadratic equation in sin x.
#D = d^2 = b^2 - 4ac = 9 + 16 = 25# --> #d = +- 5#
#sin x = -3/4 +- 5/4#
2 solutions --> sin x = 1/2 and sin x = - 2 (rejected as < -1).
#sin x = 1/2# --> #x = pi/6# and #x = (5pi)/6# (unit circle)

Science

Math

Humanities

... and beyond

How do you find all the solutions in the interval [0, 2π) without using a calculator of #3sinx = 2cos^2x#?

1 Answer

Explanation:

Related questions

Impact of this question