Asked by Carson
Find all solutions of the equation 2sin^2 x + 3sinx+1 = 0
over the interval [0,2pi)
over the interval [0,2pi)
Answers
Answered by
John
(2sinx +1)(sinx +1) = 0
2sinx + 1 = 0 sinx + 1 = 0
sinx = -1/2 sinx = -1
Can you find x? There are two possible values for the first part and one possible value for the second part given the interval.
2sinx + 1 = 0 sinx + 1 = 0
sinx = -1/2 sinx = -1
Can you find x? There are two possible values for the first part and one possible value for the second part given the interval.
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