The number of values of x in the interval [0, 3π] satisfying the equa

1.	The number of values of x in the interval [0, 3π] satisfying the equation 2 sin2 x + 5 sin x - 3 = 0 is
Answer» $\implies$ GIVEN, x + 5 sin x – 3 = 0 2 sin x(sin x + 3) – 1(sin x + 3) = 0 (2 sin x – 1)(sin x + 3) = 0 sin x = 1/2, sin x = -3 {not possible SINCE -1 ≤ sin x ≤ 1} Thus, x = π/6, 5π/6, 13π/6, 17π/6 Hence, the number of values of x in the INTERVAL [0, 3π] are 4. hence the answer is 4

The number of values of x in the interval [0, 3π] satisfying the equation 2 sin2 x + 5 sin x - 3 = 0 is

Answer»

$\implies$

GIVEN,

x + 5 sin x – 3 = 0

2 sin x(sin x + 3) – 1(sin x + 3) = 0
(2 sin x – 1)(sin x + 3) = 0
sin x = 1/2, sin x = -3 {not possible SINCE -1 ≤ sin x ≤ 1}
Thus, x = π/6, 5π/6, 13π/6, 17π/6
Hence, the number of values of x in the INTERVAL [0, 3π] are 4.

hence the answer is 4

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The number of values of x in the interval [0, 3π] satisfying the equation 2 sin2 x + 5 sin x - 3 = 0 is

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