How do you solve #3(1-sinx)=2cos^2x# in the interval #0<=x<=2pi#?

Question

4880 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-09T22:36:40

#3 - 3sinx = 2cos^2x#

Apply the identity #sin^2theta + cos^2theta = 1- > cos^2theta = 1 - sin^2theta#.

#3 - 3sinx = 2(1 - sin^2x)#

#3 - 3sinx = 2 - 2sin^2x#

#2sin^2x - 3sinx + 1 = 0#

#2sin^2x - 2sinx - sinx + 1 = 0#

#2sinx(sinx - 1) - (sinx - 1) = 0#

#(2sinx - 1)(sinx - 1) = 0#

#sinx = 1/2 and sinx = 1#

#x = pi/6, (5pi)/6, pi/2#

Hopefully this helps!