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)

## Answer this Question

## Related Questions

- Maths - Could someone help me with this question Write in polar form : sinx - i ...
- Math - What is the derivative of ((e^pie/3)(cosx))? Is it pie/3e^pie/3 times ...
- Pre-Calc - Trigonometric Identities Prove: (tanx + secx -1)/(tanx - secx + 1)= ...
- Maths - Solve this equation fo rx in the interval 0<=x<=360 3sinxtanx=8 I ...
- Math - How many solutions does the equation cosx + 1/2 = 1 have for 0<x<...
- Calculus grade 12 - solve for (<_ = less than or equal to / pie = pie sign...
- Calculus 2 Trigonometric Substitution - I'm working this problem: ∫ [1-tan...
- advanced functions grd 12 - im not sure wheer to start and how to do solve this ...
- calculus - Can someone explain how find the average rate change for these ...
- Math - Verify the identity . (cscX-cotX)^2=1-cosX/1+cosX _______ sorry i cant ...

More Related Questions