Trigonometric EquationHard
Question
The number of values of x in the interval [0,5π] satisfying the equation [0,5π] is
Options
A.0
B.5
C.6
D.10
Solution
Given, 3sin2 x - 7sin x + 2 = 0
⇒ 3sin2 x - 6sin x + 2 = 0
⇒ 3sin x(sin x - 2) - 1(sin x - 2) = 0
⇒ (3sin x - 1)(sin x - 2) = 0
⇒ sin x =
(∵ sin x = 2 is rejecterd)
⇒ x = nπ + (-1)n sin-1, n ∈ I
For 0 ≤ n ≤ 5x, ∈ [0, 5π]
There are six values of x ∈ [0, 5π] which satisfy the equation
3sin2 x - 7sin x + 2 = 0
⇒ 3sin2 x - 6sin x + 2 = 0
⇒ 3sin x(sin x - 2) - 1(sin x - 2) = 0
⇒ (3sin x - 1)(sin x - 2) = 0
⇒ sin x =
(∵ sin x = 2 is rejecterd)⇒ x = nπ + (-1)n sin-1
, n ∈ I For 0 ≤ n ≤ 5x, ∈ [0, 5π]
There are six values of x ∈ [0, 5π] which satisfy the equation
3sin2 x - 7sin x + 2 = 0
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