How do you solve #9 cos^3 x − 63 cos^2 x + cos x − 7 = 0 #?

1 Answer
Oct 25, 2015

No solution.

Explanation:

#[1]" "9cos^3x-63cos^2x+cosx-7=0#

Pair #9cos^3x# and #cosx#. Pair #-63cos^2x# and #-7#.

#[2]" "(9cos^3x+cosx)+(-63cos^2x-7)=0#

Factor out each group.

#[3]" "(cosx)(9cos^2x+1)+(-7)(9cos^2x+1)=0#

Factor out #9cos^2x+1# from the entire expression.

#[4]" "(9cos^2x+1)(cosx-7)=0#

For this to be true, #(9cos^2x+1)# or #(cosx-7)# must be equal to #0#. However, #(cosx-7)# will never be #0# because the range of the cosine function is only #[-1,1]#. Therefore, we can't use that to find #x#.

#[5]" "9cos^2x+1=0#

#[6]" "9cos^2x=-1#

#[7]" "cos^2x=-1/9#

However, this doesn't make any sense because any real number that is squared must be positive. #cos^2x# can not be equal to #-1/9#.

Therefore, we can conclude that the equation has no solution.