Solving Trigonometric Equations with Identities

Trigonometric equations are equations that contain trigonometric functions like sinθ, cosθ, or tanθ as the unknowns. They are similar to algebraic equations but use trigonometric ratios instead of variables. These equations have multiple solutions because trigonometric functions repeat their values after a fixed interval (usually 2π).

Check:

• Trigonometric Identities
• Trigonometric Equations
• Trigonometric Functions

Solved Questions

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.

• If sin⁡(x) = 0, then x = 0, π
• If sin⁡(x)= 1, then x = π/2​

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

Example 2: 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​.

Example 3: 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}​.

Example 4: \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​.

Example 5: 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​.

Example 6: 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​.

Example 7: 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​.

Example 8: 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​.

Example 9: 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}}).

Example 10: 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

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

Q2. Solve \cos(x) - \frac{\sqrt{3}}{2} = 00 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.