Posted by **Josh** on Monday, October 10, 2011 at 9:32am.

solve this equation on the given interval expressing the solution for x in terms of inverse trigonometric functions.

6(cosx)^2-cosx-5=0 on (pie/2, pie)

- Calculus -
**bobpursley**, Monday, October 10, 2011 at 9:37am
Isn't this a quadratic in cosx ?

6y^2-y-5=0

y=(1+-sqrt(1+120))/12

where y= cos x

then solve for x

- Calculus -
**Reiny**, Monday, October 10, 2011 at 9:43am
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)

- bobpursley is right - Calculus -
**Reiny**, Monday, October 10, 2011 at 9:47am
I misread your equation, should be

x = cos^-1 ( 1 ± √121)/12)

= cos^-1 (-5/6)

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