Asked by Sara
4cos2x - cosx-3 =0
0 to 2pi
0 to 2pi
Answers
Answered by
Reiny
recall cos 2x = 2cos^2 x - 1
4cos2x - cosx-3 =0
4(2cos^2 x - 1 ) - cosx - 3 = 0
8cos^2x - 4 - cosx - 3 = 0
8cos^2 x - cosx - 7 = 0
(cosx - 1)(8cosx + 7) = 0
cosx = 1 or cosx = -7/8
cosx = 1 ---> x = 0 or x = 2π
cosx = -7/8 ---> x = π - cos^-1 (7/8) or x = π + cos^-1 (7/8)
x = appr 2.64 or x = 3.64
let's check one of them
x = 3.64
LS = 4cos(7.28) - cos(3.64) - 3
= .027.. , looks good, close to zero
4cos2x - cosx-3 =0
4(2cos^2 x - 1 ) - cosx - 3 = 0
8cos^2x - 4 - cosx - 3 = 0
8cos^2 x - cosx - 7 = 0
(cosx - 1)(8cosx + 7) = 0
cosx = 1 or cosx = -7/8
cosx = 1 ---> x = 0 or x = 2π
cosx = -7/8 ---> x = π - cos^-1 (7/8) or x = π + cos^-1 (7/8)
x = appr 2.64 or x = 3.64
let's check one of them
x = 3.64
LS = 4cos(7.28) - cos(3.64) - 3
= .027.. , looks good, close to zero
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