Trigonometric Equations — Chapter Test | Class 11 IMO

Q1

Find the general solution of sin 2x = sin x (where n ∈ Z).

sin 2x − sin x = 0 → 2sin x cos x − sin x = 0 → sin x (2cos x − 1) = 0. sin x = 0 → x = nπ. cos x = 1/2 → x = 2nπ ± π/3.

Q2

Find the general solution of sec²x = 4/3 (where n ∈ Z).

sec²x = 4/3 implies cos²x = 3/4 = cos²(π/6). The solution is x = nπ ± π/6.

Q3

The equation sin x = 2 has:

The range of sin x is from −1 to 1.

Q4

The number of solutions of cos 2x = 1 in the interval [0, 2π] is:

cos 2x = 1 means 2x = 2nπ → x = nπ. For x ∈ [0, 2π], the solutions are 0, π, and 2π. Thus, there are 3 solutions.

Q5

The equation cos x = −2 has:

cos x cannot be less than −1.

Q6

A rotating fan blade position satisfies cos θ = 0. Which angle can represent a principal solution?

cos θ = 0 at 90° and 270°.

Q7

Find the general solution of cos x = √3/2 (where n ∈ Z).

cos x = cos(π/6). The general solution for cos x = cos α is x = 2nπ ± α. Thus, x = 2nπ ± π/6.

Q8

The daily profit of a shopkeeper in a Chennai market fluctuates according to cos(πd/30) = 0. Find the first day d > 0 this happens.

cos(πd/30) = 0 → πd/30 = π/2 → d/30 = 1/2 → d = 15.

Q9

Find the general solution of cos(x/2) = 0 (where n ∈ Z).

cos θ = 0 implies θ = (2n + 1)π/2. So x/2 = (2n + 1)π/2 → x = (2n + 1)π.

Q10

The least positive solution of sin x = √3/2 is:

The smallest angle with sine √3/2 is 60°.

Q11

How many solutions does the equation sin x = 1/2 have in the interval [0, 2π]?

sin x is positive in the 1st and 2nd quadrants. The solutions in [0, 2π] are π/6 and 5π/6. Hence, 2 solutions.

Q12

Aarav is flying a kite in Jaipur. The string makes an angle θ with the ground such that 2sin²θ = 1. If θ is acute, what is the angle?

2sin²θ = 1 → sin²θ = 1/2. For an acute angle, sin θ = 1/√2, which corresponds to 45°.

Q13

The principal solutions of sin x = cos x are:

tan x = 1 gives 45° and 225°.

Q14

The principal solutions of cos 2x = 1/2 are:

2x = 60°,300°,420°,660°. Hence x = 30°,150°,210°,330°? Re-evaluating gives principal solutions 30°,150°,210°,330°.

Q15

The principal solutions of sin 2x = 0 for 0° ≤ x < 360° are:

2x = n×180°. Hence x = n×90° within the interval.