Do you know which equations are called Trigonometric Equations? Well, the equations which involve trigonometric functions like sin, cos, tan, cot, sec etc. are called trigonometric equations. In this article, we will look at the different solutions of trigonometric equations in detail.

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We already know that the values of \( \sin {x} \) and \( \cos {x} \) repeat after an interval of 2π. Also, the values of \( \tan {x} \) repeat after an interval of π. If the equation involves a variable 0 ≤ x < 2π, then the solutions are called principal solutions. A general solution is one which involves the integer ‘n’ and gives all solutions of a trigonometric equation. Also, the character ‘Z’ is used to denote the set of integers.

## Principal Solutions of Trigonometric Equations

Let’s look at these examples to help us understand the principal solutions:

### Example 1

Find the principal solutions of the equation \( \sin {x} \) = \( \frac {\sqrt {3}}{2} \).

Solution: We know that, \( \sin {\frac {π}{3}} \) = \( \frac {\sqrt {3}}{2} \)

Also, \( \sin {\frac {2π}{3}} \) = \( \sin (π – {\frac {π}{3}}) \)

Now, we know that \( \sin (π – x) \) = \( \sin {x} \). Hence,

\( \sin {\frac {2π}{3}} \) = \( \sin {\frac {π}{3}} \) = \( \frac {\sqrt {3}}{2} \)

Therefore, the principal solutions of \( \sin {x} \) = \( \frac {\sqrt {3}}{2} \) are x = \( \frac {π}{3} \) and \( \frac {2π}{3} \).

**Browse more Topics Under Trigonometric Functions**

- Measurement of Angles
- Introduction to Trigonometric Functions
- Domain and Range of Trigonometric functions
- Compound Angles
- Trigonometric Equations

### Example 2

Find the principal solutions of the equation \( \tan {x} \) = -\( \frac {1}{\sqrt {3}} \).

Solution. We know that,

\( \tan {\frac {π}{6}} \) = \( \frac {1}{\sqrt {3}} \)

Also, we know that \( \tan {(π – x)} \) = – \( \tan {x} \). Therefore,

\( \tan (π – \frac {π}{6}) \) = – \( \tan {\frac {π}{6}} \) = -\( \frac {1}{\sqrt {3}} \).

Further, \( \tan {(2π – x)} \) = – \( \tan {x} \). Therefore,

\( \tan (2π – \frac {π}{6}) \) = – \( \tan {\frac {π}{6}} \) = -\( \frac {1}{\sqrt {3}} \).

Hence, the principal solutions of \( \tan {x} \) = -\( \frac {1}{\sqrt {3}} \) are:

- \( \tan (π – \frac {π}{6}) \) or \( \tan {\frac {5π}{6}} \) AND
- \( \tan (2π – \frac {π}{6}) \) or \( \tan {\frac {11π}{6}} \)

## General Solutions of Trigonometric Equations

We have learnt that,

- \( \sin {x} \) = o implies x = nπ, where n ∈ Ζ
- \( \cos {x} \) = 0 implies x = (2n + 1)\( \frac {π}{2} \), where n ∈ Ζ

Let’s look at the following theorems now:

### Theorem 1: For any real numbers x and y, \(\sin {x}\) = \(\sin {y}\) implies x = nπ + (-1)^{n}y, where n ∈ Z.

**Proof:** We have, \( \sin {x} \) = \( \sin {y} \)

⇒ \( \sin {x} \) – \( \sin {y} \) = 0

Using the sum-to-product formula of trigonometric identities, we get

\( \sin {x} \) – \( \sin {y} \) = 2\( \cos {\frac {x + y}{2}} \)\( \sin {\frac {x – y}{2}} \) = 0

Hence, either \( \cos {\frac {x + y}{2}} \) = 0

Or \( \sin {\frac {x – y}{2}} \) = 0

**Scenario 1:** \( \cos {\frac {x + y}{2}} \) = 0

Since, \( \cos {\frac {x + y}{2}} \) = 0, we have

\( \frac {x + y}{2} \) = (2n + 1)\( \frac {π}{2} \) … where where n ∈ Ζ

⇒ x + y = (2n + 1)π

⇒ x = (2n + 1)π – y

⇒ x = (2n + 1)π + (-1)^{2n + 1}y

**Scenario 2:** \( \sin {\frac {x – y}{2}} \) = 0

Since, \( \sin {\frac {x – y}{2}} \) = 0, we have

\( \frac {x – y}{2} \) = nπ … where n ∈ Ζ

⇒ x – y = 2nπ

⇒ x = 2nπ + y

⇒ x = 2nπ + (-1)^{2n}y

Combining scenarios 1 and 2, we get

x = nπ + (-1)^{n}y … where n ∈ Ζ

### Theorem 2: For any real numbers x and y, \( \cos {x} \) = \( \cos {y} \) implies x = 2nπ ± y, where n ∈ Z.

**Proof: **We have, \( \cos {x} \) = \( \cos {y} \)

⇒ \( \cos {x} \) – \( \cos {y} \) = 0

Using the sum-to-product formula of trigonometric identities, we get

\( \cos {x} \) – \( \cos {y} \) = -2\( \sin {\frac {x + y}{2}} \)\( \sin {\frac {x – y}{2}} \) = 0

Hence, either \( \sin {\frac {x + y}{2}} \) = 0

Or \( \sin {\frac {x – y}{2}} \) = 0

**Scenario 1:** \( \sin {\frac {x + y}{2}} \) = 0

Since, \( \sin {\frac {x + y}{2}} \) = 0, we have

\( \frac {x + y}{2} \) = nπ … where n ∈ Ζ

⇒ x + y = 2nπ

⇒ x = 2nπ – y

**Scenario 2:** \( \sin {\frac {x – y}{2}} \) = 0

Since, \( \sin {\frac {x – y}{2}} \) = 0, we have

\( \frac {x – y}{2} \) = nπ … where n ∈ Ζ

⇒ x – y = 2nπ

⇒ x = 2nπ + y

Combining scenarios 1 and 2, we get

x = 2nπ ± y, where n ∈ Z.

### Theorem 3: If x and y are not odd-multiples of \( \frac {π}{2} \), then \( \tan {x} \) = \( \tan {y} \) implies x = nπ + y, where n ∈ Z.

**Proof: **We have, \( \tan {x} \) = \( \tan {y} \)

⇒ \( \tan {x} \) – \( \tan {y} \) = 0

⇒ \( \frac {\sin {x}}{\cos {x}} \) – \( \frac {\sin {y}}{\cos {y}} \) = 0

⇒ \( \frac {\sin {x}\cos {y} – \cos {x}\sin{y}}{\cos {x} \cos{y}} \) = 0

⇒ \( \sin {x}\cos {y} – \cos {x}\sin{y} \) = 0

Using the sum and difference formula for trigonometric identities, we get

\( \sin {x}\cos {y} – \cos {x}\sin{y} \) = \( \sin {(x – y)} \) = 0

Therefore, we have x – y = nπ … where n ∈ Z

⇒ x = nπ + y.

### Example 3

Find the solution of \( \sin {x} \) = -\( \frac {\sqrt {3}}{2} \).

Solution: We know that \( \sin {\frac {π}{3}} \) = \( \frac {\sqrt {3}}{2} \). Therefore,

\( \sin {x} \) = -\( \frac {\sqrt {3}}{2} \) = -\( \sin {\frac {π}{3}} \)

Using the unit circle properties,we get

\( \sin {x} \) = -\( \sin {\frac {π}{3}} \) = \( \sin (π + \frac {π}{3}) \) = \( \sin {\frac {4π}{3}} \)

Hence, \( \sin {x} \) = \( \sin {\frac {4π}{3}} \)

Using Theorem 1, we get, x = nπ + (-1)^{n}(\( \frac {4π}{3} \))

### Example 4

Find the solution of \( \cos {x} \) = \( \frac {1}{2} \).

Solution: We know that \( \cos {\frac {π}{3}} \) = \( \frac {1}{2} \). Therefore,

\( \cos {x} \) = \( \cos {\frac {π}{3}} \)

Using Theorem 2, we get x = 2nπ ± \( \frac {π}{3} \)

### Example 5

Find the solution of \( \tan {2x} \) = -\( \cot (x + \frac {π}{3}) \).

Solution: We know that, \( \tan ({\frac {π}{2} + x}) \) = -\( \cot {x} \). Therefore,

\( \tan {2x} \) = -\( \cot (x + \frac {π}{3}) \) = \( \tan ({\frac {π}{2} + x + \frac {π}{3}}) \)

⇒ \( \tan {2x} \) = \( \tan ({x + \frac {5π}{6}}) \)

Usint Theorem 3, we get,

2x = nπ + x + \( \frac {5π}{6} \)

⇒ x = nπ + \( \frac {5π}{6} \) … where n ∈ Z.

## More Solved Problems on Trigonometric Equations

Q1. Find the principal and general solutions of \( \sec {x} \) = 2.

Answer. We know that, \( \sec {\frac {π}{3}} \) = 2

Also, \( \sec {\frac {π}{3}} \) = \( \sec (2π – {\frac {π}{3}}) \) = \( \sec {\frac {5π}{3}} \)

Therefore, the principal solutions of \( \sec {x} \) = 2 are,

- x = \( \frac {π}{3} \) and
- x = \( \frac {5π}{3} \)

Now, we know that \( \sec {x} \) = \( \frac {1}{\cos {x}} \)

Hence, \( \sec {x} \) = \( \sec {\frac {π}{3}} \) implies

\( \cos {x} \) = \( \cos {\frac {π}{3}} \)

By using the Theorem 2, we get x = 2nπ ± \( \frac {π}{3} \), where n ∈ Z. This is the general solution of \( \sec {x} \) = 2.

**Solved Questions for You**

**Question 1:** What are the three trigonometric equations?

**Answer: **The three major functions in trigonometry refer to as Sine, Cosine and Tangent.

**Question 2: What is the tangent formula?**

**Answer:** We make use of Tangent Angle Formula is in general to calculate the angle of the right triangle. Further, in a right triangle, the tangent of an angle is basically the length of the opposite side which is divided by the length of the adjoining side.

**Question 3: What is the inverse of sin?**

**Answer:** The arcsin function is the inverse of the sin function. However, sine itself will not be invertible as it is not injective, thus it is not bijective (invertible). Further, in order to attain arcsine function, one needs to limit the domain of sine to [−π2,π2]

**Question 4: Why is Secant called Secant?**

**Answer:** Secant is a term in mathematics which is derived from the Latin word secare. It refers ‘to cut’ and if we look at a circle, a secant will intersect the circle in precisely two points and a chord in the line segment which these two points determine. Thus, it is the interval on a secant whose endpoints are these points.

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