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
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)
3 answers
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)
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)
x = cos^-1 ( 1 ± √121)/12)
= cos^-1 (-5/6)
4 answers