How to Find General Solution of Trigonometric Equation

Here, you will learn what is trigonometric equation and how to find general solution of trigonometric equation with examples.

Let’s begin –

An equation involving one or more trigonometrical ratios of unknown angles is called a trigonometrical equation.

Solution of Trigonometric Equation

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

(a) Principal solution :- The solution of the trigonometric equation lying in the interval [0, \(2\pi\)).

(b) General Solution :- Since all the trigonometric functions are many one & periodic, hence there are infinite values values of 0 for which trigonometric functions have the same value. All such possible values of 0 for which the given trigonometric function is satisfied is given by a general formula. Such a general formula is called general solution of trigonometric equation.

(c) Particular Solution :- The solution of the trigonometric equation lying in the given interval.

General Solution of Trigonometric Equation

(a) If sin \(\theta\) = 0, then \(\theta\) = n\(\pi\), n \(\in\) I (set of integers)

(b) If cos \(\theta\) = 0, then \(\theta\) = (2n+1)\(\pi\over 2\), n \(\in\) I

(c) If tan \(\theta\) = 0, then \(\theta\) = n\(\pi\), n \(\in\) I

Example : Find the general solution of trigonometric equation :
(i) \(sin2\theta\) = 0
(ii) \(tan{3\theta\over 4}\) = 0

Solution :
(i) \(sin2\theta\) = 0

\(\implies\) \(2\theta\) = \(n\pi\), where n \(\in\) Z     [sin \(\theta\) = 0, then \(\theta\) = n\(\pi\)]

\(\implies\) \(\theta\) = \(n\pi\over 2\), where n \(\in\) Z

(ii) \(tan{3\theta\over 4}\) = 0

\(\implies\) \(3\theta\over 4\) = \(n\pi\), where n \(\in\) Z     [tan \(\theta\) = 0, then \(\theta\) = n\(\pi\)]

\(\implies\) \(\theta\) = \(4n\pi\over 3\), where n \(\in\) Z


(d) If sin \(\theta\) = sin \(\alpha\), then \(\theta\) = n\(\pi\) + \({(-1)}^n\alpha\), where \(\alpha\) \(\in\) [-\(\pi\over 2\), \(\pi\over 2\)], n \(\in\) I

(e) cos \(\theta\) = cos \(\alpha\), then \(\theta\) = 2n\(\pi\) \(\pm\) \(\alpha\), where \(\alpha\) \(\in\) [0,\(\pi\)], n \(\in\) I

(f) tan \(\theta\) = tan \(\alpha\), then \(\theta\) = n\(\pi\) + \(\alpha\), where \(\alpha\) \(\in\) (-\(\pi\over 2\), \(\pi\over 2\)), n \(\in\) I

Example : Find the general solution of trigonometric equation :
(i) \(sin\theta\) = \(\sqrt{3}\over 2\)
(ii) \(cos3\theta\) = \(-1\over 2\)

Solution :
(i) A value of \(\theta\) satisfying \(sin\theta\) = \(\sqrt{3}\over 2\) is \(\pi\over 3\)

\(sin\theta\) = \(\sqrt{3}\over 2\)

\(\implies\) \(sin\theta\) = \(sin\pi\over 3\) \(\implies\) \(\theta\) = n\(\pi\) + \({(-1)}^n{\pi\over 3}\),

(ii) \(cos3\theta\) = \(-1\over 2\)

\(\implies\) \(cos3\theta\) = \(cos{2\pi\over 3}\) \(\implies\) \(3\theta\) = 2n\(\pi\) \(\pm\) \(2\pi\over 3\)

\(\implies\) \(\theta\) = 2n\(\pi\over 3\) \(\pm\) \(2\pi\over 9\)


(g) If sin \(\theta\) = 1, then \(\theta\) = 2n\(\pi\) + \(\pi\over 2\) = (4n + 1)\(\pi\over 2\), n \(\in\) I

(h) If cos \(\theta\) = 1, then \(\theta\) = 2n\(\pi\), n \(\in\) I

(i) If \(sin^2\theta\) = \(sin^2\alpha\) or \(cos^2\theta\) = \(cos^2\alpha\) or \(tan^2\theta\) = \(tan^2\alpha\), then \(\theta\) = n\(\pi\) \(\pm\) \(\alpha\), n \(\in\) I

Example : Find the general solution of trigonometric equation \(7cos^2\theta\) + \(3sin^2\theta\) = 4

Solution : We have,

\(7cos^2\theta\) + \(3sin^2\theta\) = 4

\(7(1-sin^2\theta)\) + \(3sin^2\theta\) = 4

\(4sin^2\theta\) = 3

\(sin^2\theta\) = \(3\over 4\) = \(({\sqrt{3}\over 2})^2\)

\(sin^2\theta\) = \(sin^2{\pi\over 3}\) \(\implies\) \(\theta\) = \(n\pi\pm{\pi\over 3}\).

Hope you learnt how to solve general solution of trigonometric equation, learn more concepts of trigonometric equation and practice more questions to get ahead in the competition. Good luck!

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