06Ctrigeqnseg

Solving Trig Equations

More Examples


 * Example 1 **

math . \qquad \text{Solve } \sqrt{3} \tan \left( 3x + \dfrac{\pi}{4} \right) + 1 = 0 \; \text{ for the domain } \; x \in \big[ 0, \, 2\pi \big] math


 * Solution:**


 * ** First, rearrange to make tan the subject **

math . \qquad \tan \left( 3x + \dfrac{\pi}{4} \right) = -\dfrac{1}{\sqrt{3}} math


 * ** Now state the general solution: **

math . \qquad 3x + \dfrac{\pi}{4} = n\pi + \tan^{-1} \left( -\dfrac{1}{\sqrt{3}} \right) math


 * ** Using exact values: **

math . \qquad \tan^{-1} \left( -\dfrac{1}{\sqrt{3}} \right) = -\dfrac{\pi}{6},\, \dfrac{5\pi}{6},\, \text{ etc } math


 * ** {but we only need one value} **


 * ** So the general solution becomes: **

math . \qquad 3x + \dfrac{\pi}{4} = n\pi +\dfrac{5\pi}{6} math


 * ** Now solve for x on the left hand side. **

math \\ . \qquad 3x = n\pi + \dfrac{5\pi}{6} - \dfrac{\pi}{4} \\. \\ . \qquad 3x = n\pi + \dfrac{7\pi}{12} \\. \\ . \qquad \; x = \dfrac {n \pi}{3} + \dfrac{7\pi}{36} \qquad n \in Z math

This is the general solution to our equation.


 * ** To find the particular solution in the domain, [0, 2 p ] sub different values of n: **

math \\ . \qquad n = 0 \quad \Rightarrow \quad x = \dfrac{7\pi}{36} \\. \\ . \qquad n = 1 \quad \Rightarrow \quad x = \dfrac {\pi}{3} + \dfrac{7\pi}{36} = \dfrac{19\pi}{36} \\. \\ . \qquad \text{etc} math


 * ** Continue with different values of n until you go beyond the upper end of the domain. **

math . \qquad x = \left\{ \dfrac{7\pi}{36},\, \dfrac{19\pi}{36},\, \dfrac{31\pi}{36},\, \dfrac{43\pi}{36},\,\dfrac{55\pi}{36},\,\dfrac{67\pi}{36} \right\} math

Thus, this list is the solution to: math . \qquad \sqrt{3} \tan \left( 3x + \dfrac{\pi}{4} \right) + 1 = 0 \; \text{ for the domain } \; x \in [0, \, 2\pi] math as required.

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