﻿ If 2sin2θ – 5sinθ + 2 > 0, θ ∈ (0, 2π), then θ ∈: : Kaysons Education

# If 2sin2θ – 5sinθ + 2 > 0, θ ∈ (0, 2π), Then θ ∈:

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## Question

### Solution

Correct option is

#### SIMILAR QUESTIONS

Q1

The smallest positive number p for which the equation cos(p sinx) = sin(pcos x) has a solution ∈ [0, 2π]

Q2

Number of solutions of pair xy of the equation sin x sin y = min{–1, α2 – 4α + 5}, α ∈ R (where 0 < x < π, – π < y < 0) is

Q3

then sin x lies in

Q4

Number of solution of the equation

Q5

Let α β [– π π] be such that cos (α – β) = 1 and cos(α + β) = 1/e, The number of pairs of α, β satisfying the above system of equation is

Q6

The number of solution of the pair of equations 2sin2θ – cos2θ = 0 and 2cos2θ – 3sinθ = 0 in the interval [0, 2π] is

Q7

If sin 3α = 4 sin α sin (x + α) sin (x – α), then 864 sin2 x + 3620 cos2 x is equal to.

Q8

If x and y are the solution of the equation 12 sin x + 5 cos x + 5 cos x = 2y2 – 8y + 21, the value of 3864 cot (xy/2) is

Q9

The number of values of x in the interval [0, 3π] satisfying the equation 2sin2x + 5sinx – 3 = 0 is:

Q10

The number of solutions of the pair of equations

2sin2θ – cos2θ = 0

2cos2θ – 3sin θ = 0

In the interval [0, 2π] is