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Calculus

posted by on .

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 - ,

    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 - ,

    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 - ,

    I misread your equation, should be

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

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