Question 735214
Solve the equation on the given interval, expressing the solution for x in terms of inverse trigonometric functions. (Enter your answers as a comma-separated list.)
7(cos x)^2-cos x−6=0 on [π/2, π]
..  
7(cos x)^2-cos x−6=0
(7cosx+6)(cosx-1)=0
..
7cosx+6=0
7cosx=-6
cosx=-6/7
x=arccos(-6/7)
..
cosx-1=0
cosx=1
x=arccos(1)
..
solutions: x=arccos(-6/7), arccos(1)