Asked by Pâro
Solve 6sin^2 x-5cosx-2=0 in the interval 0≤x≤2π
Answers
Answered by
Reiny
6sin^2 x-5cosx-2=0
6(1 - cos^2 x) - 5cosx - 2 = 0
6 - 6cos^2 x - 5cosx - 2 = 0
6cos^2 x + 5cosx -4 = 0
(2cosx - 1)(3cosx + 4) = 0
cosx = 1/2 or cosx = -4/3, but the latter is not possible since -1 ≤ cosx < +1
x = 60° or x = 300°
in radians: x = π/3 or 5π/3
6(1 - cos^2 x) - 5cosx - 2 = 0
6 - 6cos^2 x - 5cosx - 2 = 0
6cos^2 x + 5cosx -4 = 0
(2cosx - 1)(3cosx + 4) = 0
cosx = 1/2 or cosx = -4/3, but the latter is not possible since -1 ≤ cosx < +1
x = 60° or x = 300°
in radians: x = π/3 or 5π/3
Answered by
Lin
How to get the x=60° or x=300°? I didn't catch that point.
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