How do you find the general solutions for #6cos^2x - cos x - 1 = 0#?

1 Answer
Aug 12, 2015

# { 2m pi +- pi/6, m in ZZ } uu {(2n+1) pi +- arccos(1/3), n in ZZ} #

Explanation:

To simplify matters, take # z = cos(x) #, so the equation now becomes # 6 z^2 - z - 1 = 0 #.

This can be solved by factorisation, as follows.
# 6 z^2 - z - 1 = 0 #
# => 6 z^2 - 3 z + 2 z - 1 = 0 #
# => 3 z (2 z - 1) + (2 z - 1) = 0 #
# => (2 z - 1) (3 z + 1) = 0 #
# => z = 1/2 # or # z = -1/3 #

Now, replacing # z = cos(x) #, and solving for #x#, we get:

  • # cos(x) = 1/2 => x = 2m pi +- pi/6, m in ZZ #
  • # cos(x) = -1/3 #
    # => x = (2n) pi +- arccos(-1/3), n in ZZ #
    # => x = (2n+1) pi +- arccos(1/3), n in ZZ #