01.8trigeqnegs

Solving Trig Equations

More Examples

Example 1

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

Solution:

First, rearrange to make tan() the subject
… … $\tan \left( 3x + \dfrac{\pi}{4} \right) = -\dfrac{1}{\sqrt{3}}$

Now state the general solution:

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

Using exact values:

… … $\tan^{-1} \left( -\dfrac{1}{\sqrt{3}} \right) = -\dfrac{\pi}{6},\, \dfrac{5\pi}{6},\, \text{ etc }$ … {but we only need one value}

So the general solution becomes:

… … $3x + \dfrac{\pi}{4} = n\pi +\dfrac{5\pi}{6}$

Now solve for x on the left hand side.

… … $3x = n\pi + \dfrac{5\pi}{6} - \dfrac{\pi}{4}$

… … $3x = n\pi + \dfrac{7\pi}{12}$

… … $\; x = \dfrac {n \pi}{3} + \dfrac{7\pi}{36} \qquad n \in Z$

This is the general solution to our equation.

To find the particular solution in the domain, $x \in[0, 2\pi]$ sub different values of n:

… … $n = 0 \quad \Rightarrow \quad x = \dfrac{7\pi}{36}$

… … $n = 1 \quad \Rightarrow \quad x = \dfrac {\pi}{3} + \dfrac{7\pi}{36} = \dfrac{19\pi}{36}$

… … … etc

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

… … $x = \left\{ \dfrac{7\pi}{36},\, \dfrac{19\pi}{36},\, \dfrac{31\pi}{36},\, \dfrac{43\pi}{36},\,\dfrac{55\pi}{36},\,\dfrac{67\pi}{36} \right\}$

Thus, this list is the solution to:

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

… … as required.

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