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Mathematics
Quincy Pratt
6 July, 16:10
Find the Solution of 2cos^2x + 3sin x = 0
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Jolie Russell
6 July, 16:52
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2cos^2 x + 3sin x = 0
2 (1 - sin^2 x) + 3sin x = 0
2 - 2sin^2 x + 3sin x = 0
2sin^2 x - 3sin x - 2 = 0
Let sin x = m, then
2m^2 - 3m - 2 = 0
2m^2 + m - 4m - 2 = 0
m (2m + 1) - 2 (2m + 1) = 0
(m - 2) (2m + 1) = 0
m = 2 or m = - 1/2
Now, sin x = - 1/2
Therefore, x = 1/6 (12nπ - π) and 1/6 (12nπ + 7π)
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