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Mathematics
Amiyah Nielsen
10 August, 07:45
Solve: cos2x+3sinx-2=0
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Ivan
10 August, 10:38
0
Cos (2x) = 1 - 2sin^2x
cos (2x) + 3sin (x) - 2 = 0
1 - 2sin^2 (x) + 3sin (x) - 2 = 0
-2sin^2 (x) + 3sin (x) - 1 = 0
2sin^2 (x) - 3sin (x) + 1 = 0
2sin^2 (x) - 2sin (x) - sin (x) + 1 = 0
2sin (x) (sin (x) - 1) - 1 (sin (x) - 1) = 0
(sin (x) - 1) (2sin (x) - 1) = 0
sin (x) = {1/2, 1}
In the range [0, 2π], x = {π/6, π/2, 5π/6}
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