Asked by Bontle
Determine the general solution of 8 cos^2x - 4cosx - 1 =0
(8cos^2x - 4cosx)/8 =1/8
cos^2x - 4cosx=1/8
cos^2x/cosx =1+4cosx/cosx
cosx = 5
x = cos^-1(5)
x = error
(8cos^2x - 4cosx)/8 =1/8
cos^2x - 4cosx=1/8
cos^2x/cosx =1+4cosx/cosx
cosx = 5
x = cos^-1(5)
x = error
Answers
Answered by
Steve
no, no. It's just a quadratic equation, in cosx, rather than just x
using the quadratic formula, we have
cosx = (1±√3)/4 or, -0.183, 0.683
So, just find the angles with those values as cosine.
using the quadratic formula, we have
cosx = (1±√3)/4 or, -0.183, 0.683
So, just find the angles with those values as cosine.
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