Trigonometric Equation Examples

Example 1 : Find general solution of (2sinx – cosx)(1 + cosx) = $sin^2x$

Solution : (2sinx – cosx)(1 + cosx) – (1 – $cos^2x$) = 0

$\therefore$   (1 + cosx)(2sinx – cosx – 1 + cosx) = 0

$\therefore$   (1 + cosx)(2sinx – 1) = 0

$\implies$   cosx = -1   or   sinx = $1\over 2$

$\implies$   cosx = -1 = cos$\pi$   $\implies$   x = 2n$\pi$ + $\pi$ = (2n+1)$\pi$, n $\in$ I

or   sinx = $1\over 2$ = sin$\pi\over 6$   $\implies$   x = k$\pi + (-1)^k{\pi\over 6}$, k $\in$ I

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Example 2 : Solve : 6 – 10cosx = 3$sin^2x$

Solution : we have, 6 – 10cosx = 3$sin^2x$

$\therefore$   6 – 10cosx = 3 – 3$cos^2x$

$\implies$   3$cos^2x$ – 10cosx + 3 = 0

$\implies$   (3cosx-1)(cosx-3) = 0   $\implies$   cosx = $1\over 3$ or cosx = 3

Since cosx =3 is not possible as -1 $\le$ cosx $\le$ 1

$\therefore$   cosx = $1\over 3$ = cos($cos^{-1}{1\over 3}$)   $\implies$   x = 2n$\pi$ $\pm$ $cos^{-1}{1\over 3}$

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Example 3 : Solve : cos3x + sin2x – sin4x = 0

Solution : we have, cos3x + (sin2x – sin4x) = 0

= cos3x – 2sinx.cos3x = 0

$\implies$   (cos3x)(1 – 2sinx) = 0

$\implies$   cos3x = 0   or   sinx = $1\over 2$

$\implies$   cos3x = 0 = cos$\pi\over 2$   or   sinx = $1\over 2$ = sin$\pi\over 6$

$\implies$   3x = 2n$\pi$ $\pm$ $\pi\over 2$   or   x = m$\pi$ + ${(-1)}^m$$\pi\over 6$

$\implies$   x = $2n\pi\over 3$ $\pm$ $\pi\over 6$   or   x = m$\pi$ + ${(-1)}^m$$\pi\over 6$; (n, m $\in$ I)

Practice these given trigonometric equation examples to test your knowledge on concepts of trigonometric equation.