We will learn how to find the general solution of trigonometric equation of various forms using the identities and the different properties of trig functions.

For trigonometric equation involving powers, we need to solve the equation either by using quadratic formula or by factoring.

**1.** Find the general solution of the equation 2 sin\(^{3}\) x - sin x = 1. Hence find the values between 0° and 360° satisfying the given equation.

**Solution:**

Since the given equation is a quadratic in sin x, we can solve for sin x either by factorization or by using quadratic formula.

Now, 2 sin\(^{3}\) x - sin x = 1

⇒ 2 sin\(^{3}\) x - sin x - 1 = 0

⇒ 2 sin\(^{3}\) x - 2sin x + sin x - 1 = 0

⇒ 2 sin x (sin x - 1) + 1 (sin x - 1) = 0

⇒ (2 sin x + 1)(sin x - 1) = 0

⇒ Either, 2 sin x + 1 = 0 or, sin x - 1 = 0

⇒ sin x = -1/2 or sin x = 1

⇒ sin x = \(\frac{7π}{6}\) or sin x = \(\frac{π}{2}\)

⇒ x = nπ + (-1)\(^{n}\)\(\frac{7π}{6}\) or x = nπ + (-1)\(^{n}\)\(\frac{π}{2}\), where n = 0, ± 1, ± 2, ± 3, …….

⇒ x = nπ + (-1)\(^{n}\)\(\frac{7π}{6}\) ⇒ x = …….., \(\frac{π}{6}\), \(\frac{7π}{6}\), \(\frac{11π}{6}\), \(\frac{19π}{6}\), …….. or x = nπ + (-1)\(^{n}\)\(\frac{π}{2}\) ⇒ x = …….., \(\frac{π}{2}\), \(\frac{5π}{2}\), ……..

Therefore the solution of the given equation between 0° and 360° are \(\frac{π}{2}\), \(\frac{7π}{6}\), \(\frac{11π}{6}\) i.e., 90°, 210°, 330°.

**2.** Solve the trigonometric equation sin\(^{3}\)
x + cos\(^{3}\) x = 0 where 0° < x < 360°

**Solution:**

sin\(^{3}\) x + cos\(^{3}\) x = 0

⇒ tan\(^{3}\) x + 1 = 0, dividing both sides by cos x

⇒ tan\(^{3}\) x + 1\(^{3}\) = 0

⇒ (tan x + 1) (tan\(^{2}\) x - tan x + 1) = 0

Therefore, either, tan x + 1 = 0 ………. (i) or, tan\(^{2}\) x - tan θ + 1 = 0 ………. (ii)

From (i) we get,

tan x = -1

⇒ tan x = tan (-\(\frac{π}{4}\))

⇒ x = nπ - \(\frac{π}{4}\)

From (ii) we get,

tan\(^{2}\) x - tan θ + 1 = 0

⇒ tan x = \(\frac{1 \pm \sqrt{1 - 4\cdot 1\cdot 1}}{2\cdot 1}\)

⇒ tan x = \(\frac{1 \pm \sqrt{- 3}}{2}\)

Clearly, the value of tan x, are imaginary; hence, there is no real solution of x

Therefore, the required general solution of the given equation is:

x = nπ - \(\frac{π}{4}\) …………. (iii) where, n = 0, ±1, ±2, ………………….

Now, putting n = 0 in (iii) we get, x = - 45°

Now, putting n = 1 in (iii) we get, x = π - \(\frac{π}{4}\) = 135°

Now, putting n = 2 in (iii) we get, x = π - \(\frac{π}{4}\) = 135°

Therefore, the solutions of the equation sin\(^{3}\) x + cos\(^{3}\) x = 0 in 0° < θ < 360° are x = 135°, 315°.

**3.**
Solve the equation tan\(^{2}\) x = 1/3 where, -
π ≤ x ≤ π.

** Solution:
**

tan 2x= \(\frac{1}{3}\)

⇒ tan x= ± \(\frac{1}{√3}\)

⇒ tan x = tan (±\(\frac{π}{6}\))

Therefore, x= nπ ± \(\frac{π}{6}\), where n = 0, ±1, ±2,…………

When, n = 0 then x = ± \(\frac{π}{6}\) = \(\frac{π}{6}\) or,- \(\frac{π}{6}\)

If n = 1 then x = π ± \(\frac{π}{6}\) + \(\frac{5π}{6}\) or,- \(\frac{7π}{6}\)

If n = -1 then x = - π ± \(\frac{π}{6}\) =- \(\frac{7π}{6}\), - \(\frac{5π}{6}\)

Therefore, the required solutions in – π ≤ x ≤ π are x = \(\frac{π}{6}\), \(\frac{5π}{6}\), - \(\frac{π}{6}\), - \(\frac{5π}{6}\).

**General solution of the equation sin x = ½****General solution of the equation cos x = 1/√2****G****eneral solution of the equation tan x = √3****General Solution of the Equation sin θ = 0****General Solution of the Equation cos θ = 0****General Solution of the Equation tan θ = 0****General Solution of the Equation sin θ = sin ∝****General Solution of the Equation sin θ = 1****General Solution of the Equation sin θ = -1****General Solution of the Equation cos θ = cos ∝****General Solution of the Equation cos θ = 1****General Solution of the Equation cos θ = -1****General Solution of the Equation tan θ = tan ∝****General Solution of a cos θ + b sin θ = c****Trigonometric Equation Formula****Trigonometric Equation using Formula****General solution of Trigonometric Equation****Problems on Trigonometric Equation**

**11 and 12 Grade Math**

**From General solution of Trigonometric Equation to HOME PAGE**

**Didn't find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.**

## New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.