Question

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

Answer 1

$\implies$

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

Answer 2

$\huge \mathbb \pink{ANSWER}$

Given,

$\implies$x + 5 sin x – 3 = 0

$\leadsto$2 sin x(sin x + 3) – 1(sin x + 3) = 0

$\leadsto$(2 sin x – 1)(sin x + 3) = 0

$\leadsto$sin x = 1/2, sin x = -3 {not possible since -1 ≤ sin x ≤ 1}

$\leadsto$Thus, x = π/6, 5π/6, 13π/6, 17π/6

$\leadsto$Hence, the number of values of x in the interval [0, 3π] are 4.

hence the answer is 4