Principal and General Solutions: The trigonometric equations involve trigonometric functions with angles as variables. The angle of \(x\) trigonometric functions such as \(\sin x, \cos x, \tan x\) is used as a variable in trigonometric equations. The solutions of trigonometric equations are values of angles that satisfy the trigonometric equation.

Similar to algebraic equations, trigonometric equations also have two types of solutions, such as the principal and general solutions. In this article, let us learn more about trigonometric equations, how to solve them, and find the solutions using a few solved examples to better understand the concepts.

What is a Trigonometric Equation?

The equations containing trigonometric functions of unknown angles are called trigonometric equations.

Example: \(\cos x=-\frac{1}{2}, \sin x=1, \tan x-\tan y=0\)

Solution of a Trigonometric Equation

A solution is a value of the unknown angle that satisfies the given trigonometric equation.

Example: Consider the equation \(\sin x=\frac{1}{2}\) This equation is satisfied by \(x=\frac{\pi}{6}, \frac{5 \pi}{6}\) etc.

So, these values of \(x\) are its solutions.

Since trigonometric functions are periodic, the trigonometric equation will have many solutions.

Example: The solution of the above equation \(\sin x=\frac{1}{2}\) can be written as \(x = n\pi + {( – 1)^n}\frac{\pi }{6}\), where \(n\) is an integer. This solution, with the variable \(n\) is called the general solution of the trigonometric equation \(\sin x=\frac{1}{2}\).

General Solutions

Sine, cosine, and tangent are the basic trigonometric ratios. Hence, the solutions can be derived using these three ratios for other equations. Similarly, the solutions for the other three ratios, such as secant, cosecant, and cotangent, can also be obtained.

Now, let us begin with the primary trigonometric equation \(\sin x=0\). The principal solutions of the given trigonometric equation are \(x=0, \pi, 2 \pi, …\). As these values satisfy the given trigonometric equation lying in the interval \([0,2 \pi]\), we can see that, for \(\sin x = 0,x = 0,\pi ,2\pi ,-\pi, – 2\pi , – 6\pi \) are also solutions.

The general solution of \(\sin x=0\) is given by \(x=n \pi, n \in Z\)

Proof of General Solutions

Using the image given below, let us prove general solutions of some standard trigonometric equations.

1. The general solution of \(\sin x=0\) is given by \(x=n \pi, n \in Z\)

Proof: Consider the diagram given above.

In \(\triangle O M P\), we have

\(\sin x=\frac{P M}{O P}\)

Now, \(\sin x=0\)

\(\Rightarrow \frac{P M}{O P}=0\)

\(\Rightarrow P M=0\)

\(\Rightarrow O P\) coincides with \(OX\) or \(OX’\)

\(\Rightarrow x=0, \pi, 2 \pi, \ldots,-\pi,-2 \pi,-3 \pi, \ldots\)

\(\Rightarrow x=n \pi, n \in Z\)

Hence, \(x=n \pi, n \in Z\) is the general solution of \(\sin x=0\)

2. The general solution of \(\tan x=0\) is \(x=n \pi, n \in Z\)

Proof: From the above figure, we get

In \(\triangle O M P\)

\(\tan x=\frac{P M}{O M}\)

Since \(\tan x=0\)

\(\Rightarrow \frac{P M}{O M}=0\)

\(\Rightarrow P M=0\)

\(\Rightarrow O P\) coincides with \(OX\) or \(OX’\)

\(\Rightarrow x=0, \pi, 2 \pi, \ldots .,-\pi,-2 \pi, \ldots\)

\(\Rightarrow x=n \pi, n \in Z\)

Hence, \(x=n \pi, n \in Z\) is the general solution of \(\tan x=0\)

3. The general solution of \(\cos x=0\) is \(x=(2 n+1) \frac{\pi}{2}, n \in Z\)

Proof: Consider the image given above.

Here,

\(\cos x=\frac{O M}{O P}\)

Given: \(\cos x=0\)

\(\Rightarrow \frac{O M}{O P}=0\)

\(\Rightarrow O P\) coincides with \(OY\) or \(OY’\)

\(\Rightarrow x=\pm \frac{\pi}{2}, \pm \frac{3 \pi}{2}, \pm \frac{5 \pi}{2}, \cdots\)

\(\Rightarrow x=(2 n+1) \frac{\pi}{2}, n \in Z\)

Hence, the general solution of \(\cos x=0\) is \(x=(2 n+1) \frac{\pi}{2}, n \in Z\).

4. The general solution of \(\cot x=0\) is \(x=(2 n+1) \frac{\pi}{2}, n \in Z\)

Proof: Consider the figure above.

By definition, we have,

\(\cot x=\frac{O M}{P M}\)

Given: \(\cot x=0\)

\(\Rightarrow \frac{O M}{P M}=0\)

\(\Rightarrow O P\) coincides with \(OY\) or \(OY’\)

\(\Rightarrow x=\pm \frac{\pi}{2}, \pm \frac{3 \pi}{2}, \pm \frac{5 \pi}{2}, \ldots\)

\(\Rightarrow x=(2 n+1) \frac{\pi}{2}, n \in Z\)

Hence, \(x=(2 n+1) \frac{\pi}{2}, n \in Z\) is the general solution of \(\cot x=0\)

Note: Since \(\sec x \geq 1\), or \(\sec x \leq-1\) therefore \(\sec x=0\) does not have any solution. Similarly, \(\operatorname{cosec} x=0\) has no solution.

Similarly, we can prove the general solutions of \(\sin x=\sin \alpha, \cos x=\cos \alpha\) and \(\tan x=\tan \alpha\)

5. The general solution of \(\sin x=\sin \alpha\) is given by \(x=n \pi+(-1)^{n} \alpha, n \in Z\)

Proof: Consider \(\sin x=\sin \alpha\)

\(\Rightarrow \sin x-\sin \alpha=0\)

\(\Rightarrow 2 \sin \left(\frac{x-\alpha}{2}\right) \cos \left(\frac{x+\alpha}{2}\right)=0\)

\(\Rightarrow \sin \left(\frac{x-\alpha}{2}\right)=0\) or \(\cos \left(\frac{x+\alpha}{2}\right)=0\)

\(\Rightarrow \frac{x-\alpha}{2}=m \pi\) or \(\frac{x+\alpha}{2}=(2 m+1) \frac{\pi}{2}, m \in Z\)

\(\Rightarrow x=2 m \pi+\alpha, m \in Z\) or \(x=(2 m+1) \pi-\alpha, m \in Z\)

\(\Rightarrow x=\) (any even multiple of \(\pi)+\alpha\) or \(x=\) (any odd multiple of \(\pi)-\alpha\)

\(\therefore x=n \pi+(-1)^{n} \alpha\) where \(n \in Z\)

Note: The equation \(\operatorname{cosec} x=\operatorname{cosec} \alpha\) is equivalent to \(\sin x=\sin \alpha\)

Hence, \(\operatorname{cosec} x=\operatorname{cosec} \alpha\) and \(\sin x=\sin \alpha\) have the same general solution.

6. The general solution of \(\cos x=\cos \alpha\) is given by \(x=2 n \pi \pm \alpha\), where \(n \in Z\)

Proof: Consider \(\cos x=\cos \alpha\)

\(\Rightarrow \cos x-\cos \alpha=0\)

\(\Rightarrow-2 \sin \left(\frac{x+\alpha}{2}\right) \sin \left(\frac{x-\alpha}{2}\right)=0\)

\(\Rightarrow \sin \left(\frac{x+\alpha}{2}\right)=0\) or \(\sin \left(\frac{x-\alpha}{2}\right)=0\)

\(\Rightarrow \frac{x+\alpha}{2}=n \pi\) or \(\frac{x-\alpha}{2}=n \pi\), where \(n \in Z\).

\(\Rightarrow x=2 n \pi-\alpha\) or \(x=2 n \pi+\alpha, n \in Z\)

\(\therefore x=2 n \pi \pm \alpha, n \in Z\)

Note: Since \(\sec x=\sec \alpha\) is equivalent to \(\cos x=\cos \alpha\) So, the general solution of \(\cos x=\cos \alpha\) and \(\sec x=\sec \alpha\) are same.

7. The general solution of \(\tan x=\tan \alpha\) is given by \(x=n \pi+\alpha, n \in Z\).

Proof: Consider \(\tan x=\tan \alpha\)

\(\Rightarrow \frac{\sin x}{\cos x}=\frac{\sin \alpha}{\cos \alpha}\)

\(\Rightarrow \sin x \cos \alpha-\cos x \sin \alpha=0\)

\(\Rightarrow \sin (x-\alpha)=0\)

\(\Rightarrow x-\alpha=n \pi, n \in Z\)

\(\therefore x=n \pi+\alpha, n \in Z\)

Note: Since \(\tan x=\tan \alpha\) is equivalent to \(\cot x=\cot \alpha\). So, general solutions of \(\tan x = \tan \alpha \) and \(\cot x=\cot \alpha\) are the same.

Steps to Calculate General Solutions

To find the general solutions of trigonometric equations of the form \(\sin x = \sin \alpha ,\cos x = \cos \alpha \) and \(\tan x=\tan \alpha\), we may use the following algorithm.

Step 1: Find a value of \(x\) preferably between \(0\) and \(2 \pi\) or between \(-\pi\) and \(\pi\) satisfying the given equation.

Step 2: Let’s call this value of \(x\) as \(\alpha\).

• If the equation is \(\sin x=\sin \alpha\), write \(x=n \pi+(-1)^{n} \alpha, n \in Z\) as the general solution.
• If the equation is \(\cos x=\cos \alpha\), write \(x=2 n \pi \pm \alpha, n \in Z\) as the general solution.
• If the equation is \(\tan x=\tan \alpha\), write \(x=n \pi+\alpha, n \in Z\) as the general solution.

Similarly, let us prove general solutions of some trigonometric equations of the type \(\sin ^{2} x=\sin ^{2} \alpha, \cos ^{2} x=\cos ^{2} \alpha\) and \(\tan ^{2} x=\tan ^{2} \alpha\)

Some More General Solutions

1. The general solution of \(\sin ^{2} x=\sin ^{2} \alpha\) is given by \(x=n \pi \pm \alpha, n \in Z\)

Proof: Consider the equation \(\sin ^{2} x=\sin ^{2} \alpha\)

\(\Rightarrow 2 \sin ^{2} x=2 \sin ^{2} \alpha\)

\(\Rightarrow 1-\cos 2 x=1-\cos 2 \alpha\)

\(\Rightarrow \cos 2 x=\cos 2 \alpha\)

\(\Rightarrow 2 x=2 n \pi \pm 2 \alpha, n \in Z\)

\(\therefore x=n \pi \pm \alpha, n \in Z\)

2. The general solution of \(\cos ^{2} x=\cos ^{2} \alpha\) is given by \(x=n \pi \pm \alpha, n \in Z\)

Proof: Consider the equation, \(\cos ^{2} x=\cos ^{2} \alpha\)

\(\Rightarrow 2 \cos ^{2} x=2 \cos ^{2} \alpha\)

\(\Rightarrow 1+\cos 2 x=1+\cos 2 \alpha\)

\(\Rightarrow \cos 2 x=\cos 2 \alpha\)

\(\Rightarrow 2 x=2 n \pi \pm 2 \alpha, n \in Z\)

\(\therefore x=n \pi \pm \alpha, n \in Z\)

3. The general solution of \(\tan ^{2} x=\tan ^{2} \alpha\) is given by \(x=n \pi \pm \alpha, n \in Z\)

Proof: Consider the equation, \(\tan ^{2} x=\tan ^{2} \alpha\)

\(\Rightarrow \frac{1-\tan ^{2} x}{1+\tan ^{2} x}=\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}\)

\(\Rightarrow \cos 2 x=\cos 2 \alpha\)

\(\Rightarrow 2 x=2 n \pi \pm 2 \alpha, n \in Z\)

\(\therefore x=n \pi \pm \alpha, n \in Z\)

Trigonometric Equation	General Solution
\(\sin x=0\)	\(x=n \pi, n \in Z\)
\(\tan x=0\)	\(x=n \pi, n \in Z\)
\(\cos x=0\)	\(x=(2 n+1) \frac{\pi}{2}, n \in Z\)
\(\cot x=0\)	\(x=(2 n+1) \frac{\pi}{2}, n \in Z\)
\(\sin x=\sin \alpha\)	\(x=n \pi+(-1)^{n} \alpha, n \in Z\)
\(\cos x=\cos \alpha\)	\(x=2 n \pi \pm \alpha\), \(n \in Z\)
\(\tan x=\tan \alpha\)	\(x=n \pi+\alpha, n \in Z\)
\(\sin ^{2} x=\sin ^{2} \alpha\)	\(x=n \pi \pm \alpha, n \in Z\)
\(\cos ^{2} x=\cos ^{2} \alpha\)	\(x=n \pi \pm \alpha, n \in Z\)
\(\tan ^{2} x=\tan ^{2} \alpha\)	\(x=n \pi \pm \alpha, n \in Z\)

Principal Solutions

The solution \(\theta\) of the trigonometric equation for which \(0 \leq \theta<2 \pi\) is called the principal solution.

Example: Find the principal solution for \(\sin x=\frac{1}{2}\)

Solution:

Here \(\sin \) is positive in the first and second quadrant

So, value in the first quadrant \(=30^{\circ}\)

Value in the second quadrant \(=180^{\circ}-30^{\circ}=150^{\circ}\)

So, the principal solutions are

\(x=30^{\circ}=30^{\circ} \times \frac{\pi}{180}=\frac{\pi}{6}\)

\(x=150^{\circ}=150^{\circ} \times \frac{\pi}{180}=\frac{5 \pi}{6}\)

Solved Examples on Principal and General Solutions

Q.1. Find the general solution of the trigonometric equation \(\sin \frac{3 x}{2}=0\)
Sol: Given: \(\sin \frac{3 x}{2}=0\)
\(\Rightarrow \frac{3 x}{2}=n \pi, n \in Z\) [as \(\sin x=0 \Rightarrow x=n \pi\) ]
\(\Rightarrow x=\frac{2 n \pi}{3}, n \in Z\)
Hence, the required general solution is \(x=\frac{2 n \pi}{3}, n \in Z\).

Q.2. Find the general solution of the trigonometric equation \(\sin 2x + \sin 4x + \sin 6x = 0\)
Sol: Given: \(\sin 2 x+\sin 4 x+\sin 6 x=0\)
\(\Rightarrow \sin 4 x+(\sin 2 x+\sin 6 x)=0\)
\(\Rightarrow \sin 4 x+2 \sin 4 x \cos 2 x=0\)
\(\Rightarrow \sin 4 x(1+2 \cos 2 x)=0\)
\(\Rightarrow \sin 4 x=0\) or, \(1+2 \cos 2 x=0\)
\(\therefore \sin 4 x=0\) or, \(\cos 2 x=-\frac{1}{2}\)
Now, consider, \(\sin 4 x=0\)
\(\Rightarrow 4 x=n \pi, n \in Z\)
\(\Rightarrow x=\frac{n \pi}{4}, n \in Z\)
Similarly, consider, \(\cos 2 x=-\frac{1}{2}\)
\(\Rightarrow \cos 2 x=\cos \frac{2 \pi}{3}\)
\(\Rightarrow 2 x=2 m \pi \pm \frac{2 \pi}{3}, m \in Z\)
\(\therefore x=m \pi \pm \frac{\pi}{3}, m \in Z\)
Hence, the required general solution is \(x=\frac{n \pi}{4}\) or,\(x=m \pi \pm \frac{\pi}{3}\), where \(m, n \in Z\).

Q.3. Find the general solution of the trigonometric equation \(\sin x=\frac{\sqrt{3}}{2}\)
Sol: Given: \(\sin x=\frac{\sqrt{3}}{2}\)
The value of \(x\) is satisfying \(\sin x=\frac{\sqrt{3}}{2}\) is \(\frac{\pi}{3}\)
\(\therefore \sin x=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \sin x=\sin \frac{\pi}{3}\)
\(\Rightarrow x=n \pi+(-1)^{n} \frac{\pi}{3}, n \in Z\) [as the general solution of \(\sin x=\sin \alpha\) is given by \(x = n\pi + \left. {{{( – 1)}^n}\alpha ,n \in Z} \right]\)
Hence, the required general solution is \(x=n \pi+(-1)^{n} \frac{\pi}{3}, n \in Z\)

Q.4. Find the general solution of the trigonometric equation \(\cos x + \cos 3x – 2\cos 2x = 0\)
Sol: Given \(\cos x+\cos 3 x-2 \cos 2 x=0\)
\(\Rightarrow 2 \cos 2 x \cos x-2 \cos 2 x=0\)
\(\Rightarrow 2 \cos 2 x(\cos x-1)=0\)
\(\therefore \cos 2 x=0\) or, \(\cos x-1=0\)
Now, consider \(\cos 2 x=0\)
\(\Rightarrow 2 x=(2 n+1) \frac{\pi}{2}, n \in Z\)
\(\Rightarrow x=(2 n+1) \frac{\pi}{4}, n \in Z\)
Similarly, consider \(\cos x-1=0\)
\(\Rightarrow \cos x=1\)
\(\Rightarrow \cos x=\cos 0\)
\(\Rightarrow x=2 m \pi \pm 0, m \in Z\)
\(\therefore x=2 m \pi, m \in Z\)
Hence, the required general solution is \(x=(2 n+1) \frac{\pi}{4}\) or \(x=2 m \pi\) where \(m, n \in Z\)

Q.5. Find the general solution of the trigonometric equation \(\sin 2 x+\cos x=0\)
Sol: Given: \(\sin 2 x+\cos x=0\)
\(\Rightarrow \cos x=-\sin 2 x\)
\(\Rightarrow \cos x=\cos \left(\frac{\pi}{2}+2 x\right)\)
\(\therefore x=2 n \pi \pm\left(\frac{\pi}{2}+2 x\right), n \in Z\)
Case (i): Consider the positive sign, then we have \(x=2 n \pi+\frac{\pi}{2}+2 x\)
\(\Rightarrow-x=2 n \pi+\frac{\pi}{2}, n \in Z\)
\(\Rightarrow x=-2 n \pi-\frac{\pi}{2}, n \in Z\)
\(\therefore x=2 m \pi-\frac{\pi}{2}\), where \(m=-n \in Z\)
Case (ii): Consider the negative sign \(x=2 n \pi-\left(\frac{\pi}{2}+2 x\right)\)
\(\Rightarrow 3 x=2 n \pi-\frac{\pi}{2}\)
\(\Rightarrow x=\frac{2 n \pi}{3}-\frac{\pi}{6}, n \in Z\)
Hence, the required general solution is \(x=2 m \pi-\frac{\pi}{2}\), or, \(x=\frac{2 n \pi}{3}-\frac{\pi}{6}\) where \(m, n \in Z\)

Q.6. Find the principal and general solutions of \(\sec x=2\)
Sol: Given: \(\sec x=2\)
As we know, \(\sec \frac{\pi}{3}=2\) and \(\sec \frac{5 \pi}{3}=\sec \left(2 \pi-\frac{\pi}{3}\right)=\sec \frac{\pi}{3}=2\)
Therefore, the principal solutions are \(x=\frac{\pi}{3}\) and \(\frac{5 \pi}{3}\)
Now, for a general solution, we have:
\(\cos x=\frac{1}{2}\)
\(\Rightarrow \cos x=\cos \frac{\pi}{3}\)
\(\therefore x=2 n \pi \pm \frac{\pi}{3}, n \in Z\) [as \(\cos x=\cos y \Rightarrow x=2 n \pi \pm y\), where \(n \in Z]\)

Q.7. Find the principal and general solutions of \(\tan x=\sqrt{3}\)
Sol: Given: \(\tan x=\sqrt{3}\)
As we know, \(\tan \frac{\pi}{3}=\sqrt{3}\) and \(\tan \frac{4 \pi}{3}=\tan \left(\pi+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}=\sqrt{3}\)
Therefore, the principal solutions are \(x=\frac{\pi}{3}, \frac{4 \pi}{3}\)
Now, for a general solution, we have:
\(\tan x=\tan \frac{\pi}{3}\)
\(\therefore x=n \pi+\frac{\pi}{3}\), where \(n \in Z\)

Summary

A solution to a trigonometric equation is the value of the unknown angle that satisfies the equation. The solution \(\theta\) of the trigonometric equation for which \(0 \leq \theta<2 \pi\) is called the principal solution. If the solution contains the integer \(n\), it is called the general solution. Reducing the given trigonometric equation in any of these possible standard forms helps find the values of unknown angles.

Some of the standard results are as follows: The general solution of \(\sin x=0\) and \(\tan x=0\) is given by \(x=n \pi, n \in Z\) and the general solution of \(\cos x=0\) and \(\cot x=0\) is \(x=(2 n+1) \frac{\pi}{2}, n \in Z\). The general solution of \(\sin x=\sin \alpha, \cos x=\cos \alpha, \tan x=\tan \alpha\) are \(x = n\pi + {( – 1)^n}\alpha ,n \in Z,x = 2n\pi \pm \alpha ,n \in Z\) and \(x = n\pi + \alpha ,\,n \in Z\) respectively.

Frequently Asked Questions (FAQs)

Q.1. What is the difference between general and particular solutions?
Ans: The solution \(\theta\) of the trigonometric equation for which \(0 \leq \theta<2 \pi\) is called the principal solution. If the solution contains the integer \(n\) in it is called the general solution.
For the general solution, convert the given trigonometric equation into the forms given below. Then, use their general formula to simplify.

• \(\sin x=\sin y \Rightarrow x=n \pi+(-1)^{n} y\), where \(n \in Z\)
• \(\cos x=\cos y \Rightarrow x=2 n \pi \pm y\), where \(n \in Z\)
• \(\tan x=\tan y \Rightarrow x=n \pi+y\), where \(n \in Z\)

Q.2. What is the general solution for \(\sin x=0 ?\)
Ans: The general solution to \(\sin x=0\) is given by \(x=n \pi, n \in Z\).

Q.3. What are general solutions?
Ans: A trigonometric equation will also have a general solution expressing all the values which would satisfy the given equation, and it is expressed in a generalized form in terms of \(‘n’\).

Q.4. How do you solve principal solutions?
Ans: We can use two simple steps to find the principal solution of trigonometric equations.
Step 1: Find the value of \(x\) preferably between \(0\) and \(2 \pi\) or between \(-\pi\) and \(\pi\), satisfying the given equation and call it \(\alpha\)
Step 2: Find all the values of \(\alpha\) by considering two quadrants.

Q.5. How do you find the general solution in trigonometry?
Ans: We can use two simple steps to find the general solutions to trigonometric equations.
Step 1: Find a value of \(x\) preferably between \(0\) and \(2 \pi\) or between \(-\pi\) and \(\pi\) satisfying the given equation and call it \(\alpha\)
Step 2: If the equation is \(\sin x=\sin \alpha\) write \(x=n \pi+(-1)^{n} \alpha, n \in Z\) as the general solution.
If the equation is \(\cos x=\cos \alpha\), write \(x=2 n \pi \pm \alpha, n \in Z\) as the general solution.
If the equation is \(\tan x=\tan \alpha\), write \(x=n \pi+\alpha, n \in Z\) as the general solution.

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