Trigonometric EquationHard
Question
The number of values of x in the interval [0, 3π] satisfying the equation 2sin2x + 5sinx - 3 = 0 is
Options
A.4
B.6
C.1
D.2
Solution
2 sin2x + 5 sin x - 3 = 0
⇒ (sin x + 3) (2 sin x - 1) = 0
⇒ sin x =
∴ In (0, 3π), x has 4 values
⇒ (sin x + 3) (2 sin x - 1) = 0
⇒ sin x =
∴ In (0, 3π), x has 4 values
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