Calculus
posted by Josh .
solve this equation on the given interval expressing the solution for x in terms of inverse trigonometric functions.
6(cosx)^2cosx5=0 on (pie/2, pie)

Isn't this a quadratic in cosx ?
6y^2y5=0
y=(1+sqrt(1+120))/12
where y= cos x
then solve for x 
by formula
cosx = (2 ± √124)/12
= .76129 or 1.09.. , the latter is undefined
x = 2.436 radians
or x = cos^1 ( (2√124)/12) 
I misread your equation, should be
x = cos^1 ( 1 ± √121)/12)
= cos^1 (5/6)