Solving Trigonometric Equations with Identities

Last Updated : 23 Jul, 2025

Trigonometric equations are the equations that include functions of trigonometry. These equations are used to find the significance of curves that fulfil certain requirements, and they are essential in various fields like physics, engineering, and computer illustrations.

Table of Content

What are Trigonometric Equations?

The trigonometric equations involve trigonometric functions of angles as variables. The angle of θ trigonometric functions such as Sinθ, Cosθ, and Tanθ is used as a variable in trigonometric equations. Similar to general polynomial equations, trigonometric equations also have solutions, which are referred to as principal solutions, and general solutions.

Solving a trigonometric equation applies to finding all the values of the variable that fulfil the equation within a bounded interval.

Important Trigonometric Identities

Pythagorean Identities

  • sin⁡2(x)+cos⁡2(x)=1
  • 1+tan⁡2(x)=sec⁡2(x)
  • 1+cot⁡2(x)=cosec⁡2(x)

Even Trigonometric Functions

For even functions, f(−x)=f(x). The cosine and secant functions are even:

  • cos⁡(−x)=cos⁡(x)
  • sec⁡(−x)=sec⁡(x)

Odd Trigonometric Functions

For odd functions, f(−x)=−f(x). The sine, tangent, cotangent, and cosecant functions are odd:

  • sin⁡(−x)=−sin⁡(x)
  • tan⁡(−x)=−tan⁡(x)
  • cosec⁡(−x)=−cosec⁡(x)
  • cot⁡(−x)=−cot⁡(x)

Quotient Identities

  • sin⁡(a ± b) = sin⁡(a)cos⁡(b) ± cos⁡(a)sin⁡(b)
  • cos(a ± b) = cos(a)cos(b) ∓ sin(a)sin(b)
  • \tan(a \pm b) = \frac{\tan(a) \pm \tan(b)}{1 \mp \tan(a) \tan(b)}

Double-angle formulas

  • sin(2x) = 2sin(x)cos(x)
  • cos(2x) = cos2(x) − sin2(x)
    • cos⁡(2x) = 2cos⁡2(x) − 1
    • cos⁡(2x) = 1 − 2sin⁡2(x)

Reciprocal Identities

  • tan(2x) = 1−tan2(x)
  • sec(x) = 1/cos(x)​
  • cosec⁡(x) = 1/sin⁡(x)
  • cot⁡(x) = 1/tan⁡(x)

Example 1: Solve sin⁡2(x) − sin⁡(x) = 0 for 0 ≤ x < 2π.

Solution:

Given: sin⁡2(x) − sin⁡(x) = 0.

sin⁡(x)(sin⁡(x) − 1) = 0.

Now,

sin⁡(x) = 0 or

sin⁡(x)−1=0 which simplifies to sin⁡(x)=1.

sin⁡(x)=0

Solutions: x=0, π, 2π

sin⁡(x)=1

Solution: x = π/2​.

Thus, all possible value for x are: x = 0, π/2, π, 2π.

Example 1: Solve 2sin⁡2(x) + 3sin⁡(x) − 2 = 0 for 0 ≤ x < 2π.

Solution:

Given: 2sin⁡2(x) + 3sin⁡(x) − 2 = 0.

Which is similar to quadratic equation in sin x.

Comparing with ax2 + bx + c = 0, we get

a = 2, b = 3 , c = −2.

Apply the Quadratic Formula: sin(x) = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)}

⇒ sin(x) = \frac{-3 \pm \sqrt{9 + 16}}{4}

⇒ sin(x) = \frac{-3 \pm \sqrt{25}}{4}

⇒ \sin(x) = \frac{-3 \pm 5}{4}

So, we get two possible solutions: sin⁡(x) = 2/4 = 1/2 or sin⁡(x) = −8/4 = −2

sin⁡(x) = 1/2​:

The solutions are x=π/6,5π/6.

sin⁡(x) = −2

No solution, since the sine function cannot be less than -1.

Thus, all possible value for x are: x = π/6, 5π/6​.

Trigonometric Equations: Solved problems

Problem 1: Solve \sin(x) = \frac{\sqrt{3}}{2}​​ for 0≤x<2π.

Solution:

\sin(x) = \frac{\sqrt{3}}{2}​​ corresponds to angles where sine has this value.

From the unit circle, \sin(x) = \frac{\sqrt{3}}{2}​​ at x = \frac{\pi}{3}​ and x = \frac{2\pi}{3}3}​.

Solutions: x = \frac{\pi}{3}​, x = \frac{2\pi}{3}​pi}{3}​.

Problem 2: Solve \cos(x) = -\frac{1}{2} for 0≤x<2π.

Solution:

cos⁡(x)=−1/2 corresponds to angles where cosine has this value.

From the unit circle, cos⁡(x)=−1/2​ at x=2π/3 and x=4π/3​.

Solutions: x=2π/3​, x=4π/3​.

Problem 3: Solve tan⁡(x)=1for 0≤x<2π.

Solution:

tan⁡(x)=1 corresponds to angles where tangent has this value.

From the unit circle, tan⁡(x)=1 at x=π/4​ and x=5/4.

Solutions: x=π/4, x=5π/4​.

Problem 4: Solve 2sin⁡(x)−1=0 for 0≤x<2π.

Solution:

Rearrange the equation: 2sin⁡(x)=1  ⟹  sin⁡(x)=1/2.

From the unit circle, sin⁡(x)=1/2 at x=π/6​ and x=5π/6​.

Solutions: x=π/6​, x=5π/6​.

Problem 5: Solve cos⁡2(x)=1/4​ for 0≤x<2π.

Solution:

Take the square root: cos⁡(x)=±12​.

For cos⁡(x)=1/2​: x=π/3​, x=5π/3​.

For cos⁡(x)=−1/2​: x=2π/3​, x=4π/3.

Solutions: x=π/3​, x=5π/3​, x=2π/3​, x=4π/3​.

Problem 6: Solve sin⁡(2x)=sin⁡(x) for 0≤x<2π.

Solution:

Use the double-angle identity: sin⁡(2x)=2sin⁡(x)cos⁡(x).

Set up the equation: 2sin⁡(x)cos⁡(x)=sin⁡(x).

Factor: sin⁡(x)(2cos⁡(x)−1)=0.

So, sin⁡(x)=0 or 2cos⁡(x)−1=0.

sin⁡(x)=0at x=0, x=π.

2cos⁡(x)−1=0 gives cos⁡(x)=1/2​, so x=π/3​, x=5π/3.

Solutions: x=0, x=π, x=π/3​, x=5π/3​.

Problem 7: Solve 3cos⁡2(x)−2=0 for 0≤x<2π.

Solution:

Rearrange: 3cos⁡2(x)=2  ⟹  cos⁡2(x)=2/3.

Take the square root: \cos(x) = \pm \sqrt{\frac{2}{3}}​​​​.

For \cos(x) = \sqrt{\frac{2}{3}}​​: x = \cos^{-1}(\sqrt{\frac{2}{3}}) and x = 2\pi - \cos^{-1}(\sqrt{\frac{2}{3}}).

For \cos(x) = -\sqrt{\frac{2}{3}}​​: x = \pi - \cos^{-1}(\sqrt{\frac{2}{3}}) and x = \pi + \cos^{-1}(\sqrt{\frac{2}{3}}).

Solutions: x values are x = \cos^{-1}(\sqrt{\frac{2}{3}}), x = 2\pi - \cos^{-1}(\sqrt{\frac{2}{3}}), x = \pi - \cos^{-1}(\sqrt{\frac{2}{3}}), x = \pi + \cos^{-1}(\sqrt{\frac{2}{3}}).

Problem 8: Solve tan⁡2(x)−1=0 for 0≤x<2π.

Solution:

Rearrange: tan⁡2(x)=1.

Take the square root: tan⁡(x)=±1.

For tan⁡(x)=1: x = \frac{\pi}{4}}​, x = \frac{5\pi}{4}​.

For tan⁡(x)=−1: x = \frac{3\pi}{4}​, x = \frac{7\pi}{4}​.

Solutions: x = \frac{\pi}{4}, x = \frac{5\pi}{4}​, x = \frac{3\pi}{4}​, x = \frac{7\pi}{4}.

Practice Problems: Trigonometric Equations

Q1. Solve \sin(x) + \frac{1}{2} = 0 for 0≤x<2π.

Q2. Solve \cos(x) - \frac{\sqrt{3}}{2} = 0 for 0≤x<2π.

Q3. Solve \tan(x) = \sqrt{3} for 0≤x<2π.

Q4. Solve sin⁡2(x)−cos⁡2(x)=0 for 0≤x<2π.

Q5. Solve 2sin⁡(x)cos⁡(x)=12 for 0≤x<2π.

Q6. Solve cot⁡(x)=1 for 0≤x<2π.

Q7. Solve sin⁡(2x)=sin⁡(x) for 0≤x<2π.

Q8. Solve cos⁡2(x)−sin⁡(x)=0 for 0≤x<2π.

Q9. Solve 3sin⁡2(x)−2=0 for 0≤x<2π.

Q10. Solve \tan(x) - \sqrt{3} = 0 for 0≤x<2π.

Answer Key: Trigonometric Equations

1. x = 7π/6​, 11π/6​

2. x = π/6, 11π/6​

3. x = π/3, 4π/3

4. x = π/4, 5π/4

5. x = π/6, 5π/6

6. x = π/4, 5π/4

7. x = 0, π/2, π, 2π

8. x = 0, π/6, π

9. x = π/3, 5π/3​

10. x = π/3, 4π/3​

Read More,

Comment
Article Tags:
Mathematics
School Learning
Maths
Geometry

Explore