Trigonometric Equations: Definition, Identities, General Solution, Examples

Trigonometric Equations: Trigonometry is a branch of mathematics that deals with the study of side lengths and angles included in right triangles. It is commonly used in surveying and navigation. There are six trigonometric functions commonly used. They are sine, cosine, tangent, cotangent, secant, and cosecant. The equations that involve trigonometric functions of a variable are called trigonometric equations.

In this article, we have provided all the points that are crucial to understand the concept of Trigonometry. To understand the topics thoroughly, students need to solve sample questions on them. It will help them know when and how to implement which rule of Trigonometry. Continue reading to know more about Trigonometric equations.

Applications of Trigonometry in Our Daily Life

Trigonometric Functions: Rules of Trigonometry

There are 6 core rules which form the base for all the trigonometric concepts. Students need to first build a strong foundation in these rules to understand higher concepts. The six trigonometric functions are defined as follows:

(i) Sine: \(\sin \,\theta = \frac{{{\text{opposite}}}}{{{\text{hypotenuse}}}}\)
(ii) Cosine: \(\cos \,\theta = \frac{{{\text{adjacent}}}}{{{\text{hypotenuse}}}}\)
(iii) Tangent: \(\tan \,\theta = \frac{{{\text{opposite}}}}{{{\text{adjacent}}}}\)
(iv) Cotangent: \(\cot \,\theta = \frac{{{\text{adjacent}}}}{{{\text{opposite}}}}\)
(v) Secant: \(\sec \,\theta = \frac{{hypotenuse}}{{adjacent}}\)
(vi) Cosecant: \({\rm{cosec}}\,\theta = \frac{{{\rm{hypotenuse}}}}{{{\rm{opposite}}}}\)

Trigonometric Identities

Trigonometric equations that hold true for all the values of the variables are called trigonometric identities.
Listed are the different types of trigonometric identities and examples for each.

1. Pythagorean identities
i. \(\theta + \theta = 1\)
ii. \(1 + \theta = \theta\)

2. Sum and difference identities
i. \(\sin \;\left( {u \pm v} \right)\; = \sin \;u\;\cos \;v\; + \cos \;u\;\sin \;v\)
ii. \(\tan \;\left( {u \pm v} \right)\; = \frac{{\tan \;u\;\; \pm \;\tan \;v\;}}{{1\; \mp \;\tan \;u\;\tan \;v\;}}\)

3. Sum to product identities
i. \(\sin \,u\,\cos \;\,v + \cos \,u\,\sin \,v\; = \sin \left( {u \pm v} \right)\)
ii. \(\cos \,u\; – \;\cos \,v\; = \; – 2\left( {\frac{{u + v}}{2}} \right)\;\;\sin \;\left( {\frac{{u\; – \;v}}{2}} \right)\)

4. Product to sum identities
i. \(\sin \,u\,\cos \,v\; = \;\frac{1}{2}\left[ {\sin \;\left( {u + v} \right) + \sin \;\left( {u – v} \right)} \right]\)
ii. \(\cos \,u\,\cos \,u\; = \frac{1}{2}\left[ {\cos \;\left( {u\; – v} \right)\; + \cos \left( {u + v} \right)} \right]\)

5. Reciprocal identities
i. \(\tan \,\theta = \frac{1}{{\cot \,\theta }}\)
ii. \(\cos \,\operatorname{co} \,\theta = \frac{1}{{\sec \,\theta }}\)

6. Quotient identities
i. \(\cot \,\theta = \frac{{\cos \,\theta }}{{\sin \,\theta }}\)
ii. \(\tan \,\theta = \frac{{\sin \,\theta }}{{\cos \,\theta }}\)

7. Even-odd identities
i. \(\sin \left( { – x} \right)\; = \; – \sin \,x\)
ii. \(\cos \left( { – x} \right)\; = \cos \,x\)

8. Double angle identities
i. \(\tan \,2\theta = \frac{{2\tan \,\theta }}{{1 – \theta }}\)
ii. \(\sin \,2u = 2\sin \,u\;\cos \,u\)

9. Half-angle identities
i. \(u = \frac{{1 + \cos \,2u}}{2}\)
ii. \(u = \frac{{1 – \cos \,2u}}{2}\)

10. Co-function identities
i. \(\sec \;\left( {\frac{\pi }{2} – x} \right) = x\)
ii. \(\cot \;\left( {\frac{\pi }{2} – x} \right) = \tan \,x\)

How to Solve Trigonometric Equations?

Simple equations like \(\cos \,\cos \,x = 1\) can be solved numerically, by approximation. Complex equations need a more analytic approach. To solve complex trigonometric equations, follow the five steps.

• Step 1: Rewrite the equation in terms of one function of one angle.
• Step 2: Solve for values in the trigonometric function.
• Step 3: List the various possible solutions for the angle. Solve for the angle.
• Step 4: Solve for the variable, if necessary.
• Step 5: Apply any restrictions, if available.

Let us learn the \(5\) steps involved using two example problems.
Example 1: Solve the equation: \(x – x + \sin \,x = 0\).
Example 2: Solve: \(3\left( {\frac{B}{2}} \right) – 1 = 0\).

Step 1: Change to One Function of One Angle
Similar to algebraic expressions that are made of variables, a trigonometric equation is made up of functions and angles. To solve an equation, the first step is to convert terms as one function of one angle. This means that all the terms in the equation should have the same angle and the same function. Any of the trigonometric identities can be used to make this conversion.

Example 1: Solve the equation: \(x – x +\sin \,x = 0\).
Here, observe that there are two types of functions: sine and cosine. The objective of the first step is to convert the equation into one function of one angle. Recalling the trigonometric identities, we can write \(x\) in terms of \(x\). Hence, we will get,
\(x – \left( {x – 1} \right) + \sin \,x\; = 0\)
\(x – \left( {x + 1} \right) + \sin \,x\; = 0\)
Observe that the equation is now in terms of one function of one angle. This can further be simplified in the following steps.

Example 2: Solve: \(3\left( {\frac{B}{2}} \right) – 1 = 0\)
Step 1: This is not necessary, as the equation is given in terms of one function of one angle.

Step 2: Solve for Values in the Trigonometric Function
At the end of step \(1\), the equation will have only one type of function with one type of angle. In this step, solve for the value of that function by simplification, similar to algebraic simplification. Utilise algebraic identities wherever needed.

Example 1: Continuing from the previous step, we can further simplify to find the value of the trigonometric function.
\(x – x + 1 + \sin \,x = 0\)
\(1 + \sin \,x = 0\)
\( \sin \,x = -1\)

Example 2: As Step \(1\) is not relevant to this trigonometric equation, continuing with Step \(2\).
Step 2: Let’s solve for the trigonometric function.
\(\left( {\frac{B}{2}} \right) = \frac{1}{3}\)
\(\tan \;\left( {\frac{B}{2}} \right) = \pm \sqrt {\frac{1}{3}} \)
Note: Do not miss to add the \( \pm \) sign when employing square roots during simplification.

Step 3: Solve for the Angle
After solving for the function, now, over to solve for the angle.
i. If the angles are multiples of \(\frac{\pi }{6}\) or \(\frac{\pi }{4}\), they are relatively easier to solve.
ii. If the angles are multiples of half angles, then half-angle identities are used to evaluate.
Trigonometric functions are periodic. This means that their values repeat in a cycle. Hence, there is either zero or infinitely many solutions for any trigonometric equation.
For example, consider the equation \(\tan \,\theta = 1\).
Shown here is the graph for different values of \(y = \tan \,x\).

The solutions for the equation \(\tan \,\theta = 1\) are identified from the graph as:
\(\theta = \ldots – \frac{{7\pi }}{4},\; – \frac{{3\pi }}{4},\frac{\pi }{4},\frac{{5\pi }}{4},… \)
The cycle here is \(\frac{{5\pi }}{4}\; – \;\frac{\pi }{4} = \frac{{4\pi }}{4} = \pi \)
Hence, the general solution for this function can be defined as \(\theta = \frac{\pi }{4} + n\pi \) where \(n\) is an integer.
Similarly, solve for angles in the given equation by listing out the possible solutions.

Example 1: Continuing from Step \(2\), we have to solve for \(x\).
\(\sin \,x = – 1\)
Using the general solution of trigonometric equations, we know that the general solution is,
\(x = \left( {4n – 1} \right)\frac{\pi }{2}\)
Note that not all calculations are going to be this simple.

Example 2: Continuing from Step \(2\), now we solve for angles.
Let \(\frac{B}{2} = \theta \). We have,
\(\tan \,\theta = \pm \sqrt {\frac{1}{3}} \)
That is, \(\tan \,\theta \; = \sqrt {\frac{1}{3}} \) or \(\tan \,\;\theta =\, – \sqrt {\frac{1}{3}} \)
\(\theta = \frac{\pi }{6}\) or \(\theta = \frac{5\pi }{6}\)
The general solution can be written as, \(\theta = \frac{\pi }{6} + n\pi \) or \(\theta = \frac{{5\pi }}{6} + n\pi \)
But, \(\theta = \frac{B}{2}\), hence the general solution is: \(\frac{B}{2} = \frac{\pi }{6} + n\pi \) or \(\frac{B}{2} = \frac{{5\pi }}{6} + n\pi \)

Step 4: Solve for the Variable
Solving for the angle effectively is a \(2-\)step process if there is a variable involved in the angles. In case of a double angle or a half-angle, ensure that the periodicity is added to the specific angle and not the half or double values. It is important to work with the angle and not the variable in the function of the angle.

Example 2: Continuing, we have the general solution from Step \(3\) as:
\(\frac{B}{2} = \frac{\pi }{6} + n\pi \) or \(\frac{B}{2} = \frac{5\pi }{6} + n\pi \)

Step 4: Let’s now solve for the variable.
Multiplying by \(2\) on both sides, we get,
\(B = \left( {\frac{\pi }{6} + n\pi } \right) \times 2\) or \(B = \left( {\frac{5\pi }{6} + n\pi } \right) \times 2\)
\(B = \frac{\pi }{3} + 2n\pi \) or \(B = \frac{5\pi }{3} + 2n\pi \)
Observe that the angle \(\theta\) or \(\frac {B}{2}\) has a period of \(\pi\), and the variable \(B\) has a period of \(2\pi\).

Step 5: Apply any Restriction Available
The ultimate step of solving a trigonometric equation involves applying any given restriction. The most common restriction is the interval provided. It is either provided as a notation or as an inequality.
For example:
Interval notation: \(( – \pi ,\;2\pi )\)
Inequality: \( – \pi \le \theta < 2\pi \)
Be mindful of the open or closed intervals or the slack and strict inequalities, whichever applies when provided with intervals.
In case of no restriction, Step \(4\) is the last step. The solution set will contain all real solutions as obtained in the previous step.

Example 2: Using the general solution from Step \(4\), we have:
\(B = \frac{\pi }{3} + 2n\pi \) or \(B = \frac{5\pi }{3} + 2n\pi \)
Let’s say the inequality is \(0 \le B < \pi \)

Step 5: Apply the restriction.
i. For \(n = \,- 1\),
\(B = \frac{\pi }{3} + 2\left( { – 1} \right)\pi \) or \(B = \frac{5 \pi }{3} + 2\left( { – 1} \right)\pi \)
\(B = \frac{\pi }{3} – 2\pi \) or \(B = \frac{5\pi }{3} – 2\pi \)
\(B =\, – \frac{{5\pi }}{3}\) or \(B = \,- \frac{{\pi }}{3}\)
ii. For \(n = 0\),
\(B = \frac{\pi }{3} + 2\left( { 0} \right)\pi \) or \(B = \frac{5 \pi }{3} + 2\left( { 0} \right)\pi \)
\(B = \frac{{\pi }}{3}\) or \(B = \frac{{5\pi }}{3}\)
Observe that there is just one solution that lies within the given interval.
\(\therefore \) Solution: \(B = \frac {\pi}{3}\)
This is called the principal solution of the trigonometric equation.

General Solution Vs Principal Solution

Solving trigonometric equations may result in two types of solutions – principal and general.
For trigonometric equations that contain trigonometric functions in variable \(x\), where \(x\) lies in the interval \([0,\,2\pi]\), is called the principal solution.
If the solution to a trigonometric equation is in terms of \(n\), where \(n\) is an integer, it is called a general solution.

Derivations and Proofs of Theorems

Theorem 1: For any real numbers \(x\) and \(y\) \(\sin \,x = \sin \,y\; \Rightarrow x = n\pi + {\left( { – 1} \right)^n}y,\;n \in Z\)
Proof:
Given: \(\sin \,x = \sin \,y\)
Rearranging, \(\sin \,x – \sin \,y = 0\)
\( \Rightarrow 2\cos \;\frac{{x + y}}{2}\;\sin \;\frac{{x – y}}{2} = 0\)
\(\therefore \,\cos \;\frac{{x + y}}{2} = 0\) or \(\sin \;\frac{{x – y}}{2} = 0\)
\( \Rightarrow \frac{{x + y}}{2} = \left( {2n + 1} \right)\frac{\pi }{2}\) or \(\frac{{x – y}}{2} = n\pi \), where \(n = Z\)
Simplifying, we get,
\(x = \left( {2n + 1} \right)\pi – y\) or \(x = 2n\pi + y\)
Hence, \(x = \left( {2n + 1} \right)\pi + {\left( { – 1} \right)^{2n + 1}}y\) or \(x = 2n\pi + {\left( { – 1} \right)^{2n}}y\)
Combining, we get, \(x = n\pi + {\left( { – 1} \right)^n}y,\;n \in Z\).

Theorem 2: For any real numbers \(x\) and \(y\), \(\cos \,x = \cos \,y\; \Rightarrow x = 2n\pi \pm y,\,\;n \in Z\).
Proof:
Given: \(\cos \,x = \cos \,y\)
Rearranging, \(\cos \,x – \cos \,y = 0\)
\( \Rightarrow – 2\sin \;\frac{{x + y}}{2}\;\sin \;\frac{{x – y}}{2} = 0\)
\(\therefore \,\sin \;\frac{{x + y}}{2} = 0\) or \(\sin \;\frac{{x – y}}{2} = 0\)
\( \Rightarrow \frac{{x + y}}{2} = n\pi \) or \( \frac{{x – y}}{2} = n\pi \), where \(n = Z\)
Simplifying, we get,
\(x = 2n\pi – y\) or \(x = 2n\pi + y\)
Combining, we get, \(x = 2n\pi \pm y,\,n \in Z\)

Theorem 3: If \(x\) and \(y\) are not odd multiples of \(\frac {\pi}{2}\), then prove that \(\tan \,x = \tan \,y\; \Rightarrow x = n\pi + y,\,n \in Z\).
Proof:
Given: \(\tan \,x = \tan \,y\)
Rearranging, \(\tan \,x – \tan \,y = 0\)
\( \Rightarrow \frac{{\sin \,x\;\cos \,y\; – \cos \,x\,\sin \,y}}{{\cos \,x\;\cos \,y}} = 0\)
\(\sin \,x\;\cos \,y – \cos \,x\;\sin \,y = 0\)
This can be written as,
\(\sin \,(x – y) = 0\)
\(\therefore x – y = n\pi\)
\( \Rightarrow x = n\pi + y,\,n \in Z\).

Trigonometric Equations and their General Solutions

Shown below is a list of common trigonometric equations and their general solutions, where \(n\) is an integer.

Let us now learn to solve trigonometric equations using reference angles and trigonometric identities.

Solved Problems: Trigonometric Equations

Students can check the solve questions on Trigonometric equations below and practice on their own:

Q.1. What are the solutions of \(\sin \;\left( {x\; – \;\frac{\pi }{3}} \right) =\,- \frac{1}{2}\) in the \(x-\)interval \([0,\,2\pi]\)?
Ans: Take \(x\; – \;\frac{\pi }{3} = y\)
\(\therefore \,\sin \,y\; = \,- \frac{1}{2}\)
\( \Rightarrow y = \,- \frac{\pi }{6},\,\frac{{7\pi }}{6}\) (in the interval \(\left[ {0,\,2\pi } \right] – \frac{\pi }{3} \le \left( {x – \frac{\pi }{3}} \right) \le \frac{{5\pi }}{3}\))
\(\therefore \,x =\, – \frac{\pi }{6} + \frac{\pi }{3}\) and \(x = \frac{{7\pi }}{6} + \frac{\pi }{3}\)
\( \Rightarrow x = \frac{\pi }{6},\,\frac{{3\pi }}{2}\).

Q.2. Solve \(\sin \,x\; + 2 = 3\), over the interval \({0^{\rm{o}}} \le x < {360^{\rm{o}}}.\)
Ans: \(\sin \,x\; + 2 = 3\)
\(\sin \,x = 3 – 2\)
\(\sin \,x = 1\)
General solution: \(x = \frac{{\left( {4n + 1} \right)\pi }}{2}\)
Given that: \({0^{\rm{o}}} \le x < {360^{\rm{o}}},\,x = \frac{\pi }{2}\)
\(\therefore \,x = 90^\circ\)
Note: Solution is given in degrees because inequality is provided in degrees.

Q.3. Find the exact solution of the equation: \(\alpha \; =\, -\cos \,\alpha \; + \alpha \), \({0^{\rm{o}}} \le \alpha \le {360^{\rm{o}}}.\)
Ans: Using, \(\alpha = 1 – \alpha\),
We get, \(\alpha = \,-\cos \;\alpha + \left( {1 – \alpha } \right)\)
\(\alpha \; + \alpha \; + \cos \,\alpha \; – 1 = 0\)
\(2 \alpha \; + \cos \,\alpha \; – 1 = 0\)
Factoring,
\((2 \cos \,\alpha \; – 1)(\cos\,\alpha \; + 1) = 0\)
Hence,
\((2 \cos \,\alpha \; – 1) = 0\) and \((\cos \,\alpha \; + 1) = 0\)
\(2 \cos \,\alpha = 1\) and \(\cos\,\alpha = \,- 1\)
\(\cos \,\alpha = \frac {1}{2}\)
In the interval \({0^{\rm{o}}} \le \alpha < {360^{\rm{o}}},\)
\(\alpha = 60^\circ ,\;300^\circ \) and \(\alpha = 180^\circ\)
Then, \(\alpha = 60^\circ ,\;300^\circ ,\;180^\circ\).

Q.4. Find the general solution of the equation \(\cos \;2x\; – 2\tan \;x\; + 2 = 0\)
Ans: Using \(\cos \;2x\; = \frac{{1 – x\;}}{{1 + x}}\), we get,
\(\frac{{1 – x\;}}{{1 + x\;}} – 2\tan \,x\; + 2 = 0\)
\(1 – x\; – 2\tan \,x\;\left( {1 + x\;} \right) + 2\left( {1 + x\;} \right) = 0\)
\(1 – x\; – 2\tan \,x\; – 2x\; + 2 + 2x\; = 0\)
\( – 2x\; + x\; – 2\tan \,x\; + 3 = 0\)
\(2x\; – x\; + 2\tan \,x\; – 3 = 0\)
Factoring, we get,
\(\left( {\tan \,x\; – 1} \right)\left( {2x\; + \tan \,x\; + 3} \right) = 0\)
Here,
\(\left( {2x\; + \tan \,x\; + 3} \right)\) has imaginary roots.
Therefore, \(\left( {\tan \,x\; – 1} \right) = 0\)
\( \Rightarrow \tan \,x = 1\)
\(x = \frac{\pi }{4} + n\pi \).

Q.5. Find all solutions for \(\left( {\frac{E}{2}} \right) – \;\left( {\frac{E}{2}} \right) = 1\).
Ans: Substitute \(\left( {\frac{E}{2}} \right) = 1\; – \left( {\frac{E}{2}} \right)\)
\(\left( {\frac{E}{2}} \right) – \;\left( {\frac{E}{2}} \right) = 1\)
\(\left( {\frac{E}{2}} \right) – \;1 – 1 = 0\)
\(\left( {\frac{E}{2}} \right) – \;1 = 0\)
\(\left( {\frac{E}{2}} \right) = 0\)
\(\therefore \,\frac{E}{2} = \frac{{\left( {2n + 1} \right)\pi }}{2}\)
\( \Rightarrow E = \left( {2n + 1} \right)\pi \).

Q.6. Solve \(2x – \sqrt 3 \;\cos \,x\; = 0\) in the interval \({0^{\rm{o}}} \le \alpha < {360^{\rm{o}}}.\)
Ans: \(2x – \sqrt 3 \;\cos \,x = 0\)
\(\cos \,\cos \,x\left( {\;2\cos \,x – \sqrt 3 \;} \right) = 0\)
Factoring,
\(\cos \,x = 0\) or \(2\cos \,x – \sqrt 3= 0\)
\(2\cos \,x = \sqrt 3\)
\(2\cos \,x = \frac {\sqrt 3}{2}\)
For \(\cos \,x = 0\)
\(x = 90^\circ ,\;270^\circ \)
For \(\cos \,x = \frac {\sqrt 3}{2}\),
\(x = 30^\circ ,\;2\pi – 30^\circ \)
\(x = 30^\circ ,\; 330^\circ \)
Combining the solutions, \(x = 30^\circ ,\; 90^\circ,\;270^\circ,\; 330^\circ \).

Trigonometric Equations Summary

Thus, we know what trigonometric equations are, how some primary trigonometric equations can be derived using trigonometric functions, and trigonometric identities. There are two types of solutions for trigonometric equations – general solution and principal solution. While a general solution is written in terms of n, where n is an integer, the principal solution is a trigonometric function in variable \(x\), where \(x\) lies in the interval \([0,\,2\pi]\).

Having learnt to find the solution for trigonometric equations in five steps, we have also solved a few examples using the gained knowledge.

FAQs on Trigonometric Equations

Below are some of the most frequently asked questions on Trigonometric Equations:

Now you are provided with all the necessary information on the concept of trigonometric equations and we hope this detailed article is helpful to you. Stay tuned to Embibe for more such conceptual information.