Trigonometric Equations Practice Questions (original) (raw)

Last Updated : 16 Jun, 2026

Trigonometric Equations are used to calculate the angles that meet a specific trigonometric relationship. These equations are essential in fields like physics, engineering, and navigation, where accurate angle measurements are critical.

**​Trigonometric Equations: Practice Questions with Solution

**Question 1: Solve for θ if sin (θ) = 1/2

To solve for θ given that

sin(θ) = 1/2​

We need to find the angles where the sine function equals 1/2.

Sine function equals 1/2 at specific angles within the standard range of o to 360° ( 0 to 2π radians):

sin(θ) = 1/2

**Question 2: Solve for θ if cos (θ) = -1/2.

To solve for θ given that

cos(θ) = -1/2​,

We need to find the angles where the sine function equals -1/2.

θ = arccos (-1/2) = 120° or 240°
θ = 120° + 360°(n) or 240°+ 360°(n), where n ∈ Z

**Question 3: Solve for θ if tan(θ) = 1

To solve for θ given that

tan(θ) = 1​,

we need to find the angles where the sine function equals 1.

θ = arctan (1) = 45° or 225°
θ = π/4 ​+ nπ

​**Question 4: Solve for θ: 2cos2θ - 1 = 0.

2cos2θ - 1 = 0
cos2θ = 1/2
cos θ = ± 1/√2
θ = 45o, 135o, 225o, 315o

**Question 5: Solve 2sin(θ) + 1 = 0 for 0 ≤ θ ≤ 2π.

2sin(θ) + 1 = 0
sin(θ) = -1/2
θ = 7π/6, 11π/6

**Question 6: Solve sin(θ) cos(θ) = 1/4 for 0 ≤ θ ≤ 2π.

First, use the double-angle identity for sine:

sin(θ) cos(θ) = (1/2) sin(2θ)

Thus, the equation becomes:(1/2) sin(2θ) = 1/4

Multiply both sides by 2 to solve for sin(2θ):

sin(2θ) = 1/2

Sine function has a value of 1/2 at the angles:

2θ = π/6, 5π/6, 13π/6, 17π/6
⇒ θ = π/12, 5π/12, 13π/12, 17π/12

Thus, the solution to the equation sin(θ) cos (θ) = 1/4 for 0 ≤ θ ≤ 2 are:

θ = π/12, 5π/12, 13π/12, 17π/12

**Question 7: Solve tan(2θ) = √3 for 0 ≤ θ ≤ 2π

​First, solve for 2θ in the equation tan(2θ) = √3 .

Tangent function has a value of √3 at the angles: .

2θ = π/3, 4π/3
θ = π/6, 2π/3, 7π/6, 5π/3

Thus, the solutions to the equation tan(2θ) = √3 for 0 ≤ θ ≤ 2π are:

θ = π/6, 2π,/3, 7π/6, 5π/3

**Question 8: Solve cos(3θ) = 1 for 0 ≤ θ < 2π.

First, solve for 3θ in the equation cos(3θ) = 1.

Cosine function has a value of 1 at the angles:

3θ = 0, 2π, 4π, 6π
θ = 0, 2π/3 , 4π/3

Thus, the solution to the equation cos(3θ) = 1 for 0 ≤ θ ≤ 2π are: θ = 0, 2π/3, 4π/3

Practice Problems

**Q1. Solve sin⁡(θ) = −√3/2 for 0 ≤ θ < 2π.

**Q2. Solve cos⁡(2θ) = −1 for 0 ≤ θ < 2π.

**Q3. Solve tan⁡(θ) = √3​ for 0 ≤ θ < 2π.

**Q4. Solve sin⁡(3θ) = 0 for 0 ≤ θ < 2π.

**Q5. Solve cos⁡(θ) = √2/2 for 0 ≤ θ < 2π.

**Q6. Solve sin⁡2(θ)=3/4 for 0 ≤ θ < 2π.

**Q7. Solve 3sin⁡(θ)−1 = 0 for 0 ≤ θ < 2π.

**Q8. Solve cos⁡(4θ)=1 for 0 ≤ θ < 2π.

**Q9. Solve tan⁡(3θ) = −1 for 0 ≤ θ < 2π.

**Q10. Solve sin⁡(θ)cos⁡(θ) = −1/4 for 0 ≤ θ < 2π.

Answer Key

  1. θ = 4π/3, 5π/3
  2. θ = π/4, 3π/4, 5π/4, π
  3. θ = π/3, 4π/3
  4. θ = 0, π/3, 2π/3, π, 4π/3
  5. θ = π/4, 7π/4
  6. θ = π/3, 2π/3, 4π/3, π
  7. θ = π/6, 5π/6
  8. θ = 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4
  9. θ = π/6, 7π/6, 11π/6
  10. θ = 3π/8, 7π/8, 11π/8, 15π/8