The number of values of x in the interval [0, 5π] satisfying the equa

1.	The number of values of x in the interval [0, 5π] satisfying the equation 3 sin2 x – 7sinx + 2 = 0 is(a) 0 (b) 5(c) 6 (d) 10
Answer» Answer : (c) 6 3 sin²x – 7 sin x + 2 = 0 ⇒ (3 sin x – 1) (sin x – 2) = 0 ⇒ (3 sin x – 1) = 0 or (sin x – 2) = 0 ∵ sin x = 2 is inadmissible, therefore, sin x = \(\frac{1}{3}\) Since, sin x = sin α where sin α = \(\frac{1}{3}\) , so α lies in the 1st quadrant ⇒ x = nπ + (–1)ⁿ α, n∈I, where 0 < α < π/2 Since x lies in the interval [0, 5π], so we have one value of x corresponding to each of the values 0, 1, 2, 3, 4, 5 or n. ∴ The number of values of x in the interval [0, 5π] is 6.

The number of values of x in the interval [0, 5π] satisfying the equation 3 sin2 x – 7sinx + 2 = 0 is(a) 0 (b) 5(c) 6 (d) 10

Answer»

Answer : (c) 6

3 sin²x – 7 sin x + 2 = 0

⇒ (3 sin x – 1) (sin x – 2) = 0

⇒ (3 sin x – 1) = 0 or (sin x – 2) = 0

∵ sin x = 2 is inadmissible, therefore, sin x = \(\frac{1}{3}\)

Since, sin x = sin α where sin α = \(\frac{1}{3}\) , so α lies in the 1st quadrant

⇒ x = nπ + (–1)ⁿ α, n∈I, where 0 < α < π/2

Since x lies in the interval [0, 5π], so we have one value of x corresponding to each of the values 0, 1, 2, 3, 4, 5 or n.

∴ The number of values of x in the interval [0, 5π] is 6.