How do you find the exact value of #3cos2theta-4cos^2theta+2=0# in the interval #0<=theta<360#?

1 Answer
Nov 21, 2016

Our goal here is to convert everything to one trigonometric function. Sine would probably be easiest. We hence apply the following identities:

#sin^2beta + cos^2beta = 1 -> cos^2beta = 1- sin^2beta#
#cos2beta = 1 - 2sin^2beta#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#3(1- 2sin^2theta) - 4(1 - sin^2theta) + 2 = 0#

#3 - 6sin^2theta - 4 + 4sin^2theta + 2 = 0#

#1 = 2sin^2theta#

#1/2 = sin^2theta#

#+-1/sqrt(2) = sintheta#

#theta = 45˚, 135˚, 225˚, 315˚#

Hopefully this helps!