06Hsolvingequations

= Solving Trigonometric Equations =


 * Because trig functions are cyclic (repeating over and over to infinity)
 * Any simple trig function will have infinite solutions
 * or none at all ( eg sin(q ) = 2 has no solutions )


 * Example 1 **

math . \qquad \text{Solve } \; \sin \big( \theta \big) = \dfrac{1}{2} math


 * Solution:**


 * ** Graphically, this is the set of points where y = sin( q ) intersects with y = ½ **

... ...


 * ** sin( q ) is positive in Q1 and Q2 so the first two solutions are in Q1 and Q2 **

math . \qquad \theta = \sin^{-1} \left( \dfrac{1}{2} \right) \\.\\ . \qquad \theta = \dfrac{\pi}{6} \quad and \quad \theta = \pi - \dfrac{\pi}{6} \\.\\ . \qquad \theta = \dfrac{\pi}{6} \quad and \quad \theta = \dfrac{5\pi}{6} math


 * ** Because the period of sin( q ) is 2 p, **
 * ** We can obtain the other solutions by repeatedly adding 2 p to our first two solutions **

math \textbf{Q1: } \quad \left\{ \dots,\, -2\pi+\dfrac{\pi}{6},\, \dfrac{\pi}{6},\, 2\pi+\dfrac{\pi}{6},\, 4\pi+\dfrac{\pi}{6},\, \dots \right\} = \left\{ \dots,\, -\dfrac{11\pi}{6},\, \dfrac{\pi}{6},\,\dfrac{13\pi}{6},\, \dfrac{25\pi}{6},\, \dots \right\} math

... ... ... ** and **

math \textbf{Q2: } \; \left\{ \dots,\, -2\pi+\dfrac{5\pi}{6},\, \dfrac{5\pi}{6},\, 2\pi+\dfrac{5\pi}{6},\, 4\pi+\dfrac{5\pi}{6},\, \dots \right\} = \left\{ \dots,\, -\dfrac{7\pi}{6},\, \dfrac{5\pi}{6},\,\dfrac{17\pi}{6},\, \dfrac{29\pi}{6},\, \dots \right\} math


 * T ** hese solutions can be written as: **

math . \qquad \theta = \left\{ 2n\pi + \dfrac{\pi}{6},\, n \in Z \right\} \;\; \text{ and } \;\; \theta = \left\{ \Big(2n+1 \Big) \pi - \dfrac{\pi}{6},\, n \in Z \right\} math

... ... ** Or **

math . \qquad \theta = \left\{ 2n\pi + \dfrac{\pi}{6},\, \Big(2n+1 \Big) \pi - \dfrac{\pi}{6},\, n \in Z \right\} math


 * These are called the ** general solutions ** to the equation given in Example 1


 * IMPORTANT NOTE: **
 * **Z** is the set of all integers (positive and negative).
 * Some texts use **J** for the set of integers.
 * Both **J** and **Z** are accepted.


 * General Solutions **


 * For any values of ** a ** where the function is defined,
 * The ** general solutions ** will be given by these rules:




 * Don't forget to select the correct quadrant (due to positive/negative values for a)


 * Solutions within a specified domain **


 * Often we will be asked for all of the solutions within a domain
 * If we have a general solution,
 * substitute various values of n into the rule to get a list
 * If we don't have a general solution,
 * start with the first two solutions, repeatedly add the period of the function to those two values to get a list


 * IMPORTANT NOTE: **
 * If the domain is specified in degrees, the answers should be written in degrees
 * If the domain is specified in radians, the answers should be written in radians


 * Example 2 **

math . \qquad \text{Solve } \; 3\sqrt{2}\sin \big( 4x \big) + 3 = 0 \\.\\ . \qquad \text{in the domain } \; x \in \big[ 0, \; \pi \big] math


 * Solution:**


 * ** Rearrange to make the trig function the subject **

math . \qquad 3\sqrt{2}\sin \big( 4x \big) + 3 = 0 \\.\\ . \qquad 3 \sqrt{2} \sin \big( 4x \big) = -3 \\.\\ . \qquad \sin \big( 4x \big) = \dfrac{-3}{3\sqrt{2}} \\.\\ . \qquad \sin \big( 4x \big) = \dfrac{-1}{\sqrt{2}} math


 * ** We recognise an exact value, in quadrant 1: **

math . \qquad \qquad \sin^{-1} \left( \dfrac{1}{\sqrt{2}} \right) = \dfrac{\pi}{4} math


 * ** Sin is negative in Q3 and Q4, so **

math . \qquad 4x = \sin^{-1} \left( \dfrac{-1}{\sqrt{2}} \right) \\.\\ . \qquad 4x = \pi + \dfrac{\pi}{4} \quad and \quad 2\pi - \dfrac{\pi}{4} \\.\\ . \qquad 4x = \dfrac{5\pi}{4} \quad and \quad \dfrac{7\pi}{4} math


 * ** Notice the 4x on the left is from the sin(4x) in the equation **
 * ** Now divide both answers by 4 to make x the subject **

math . \qquad x = \dfrac{5\pi}{16} \quad and \quad \dfrac{7\pi}{16} math


 * **These are our first two solutions.**

math . \qquad \text{Period } = \dfrac{2\pi}{n} \\.\\ . \qquad \text{Period } = \dfrac{2\pi}{4} = \dfrac{\pi}{2} math


 * ** Find the remaining solutions by repeatedly adding the period ( p /2) to the first two solutions **
 * ** Continue until we reach the top of the domain ** ... ** [0, p ] **

math . \qquad x = \dfrac{5\pi}{16}, \; \dfrac{7\pi}{16}, \; \dfrac{13\pi}{16}, \; \dfrac{15\pi}{16} math

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