06Ctrigeqns

toc = Trigonometric Equations =


 * Because trig functions are cyclic (repeating over and over to infinity)
 * any simple trig equation will have an infinite number of solutions.

** Example 1 **
math . \qquad \text{Solve } \sin(\theta) = \frac{1}{2} math

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


 * ** sin( **q ** ) is positive in 1st and 2nd Quadrants (Q1 and Q2) **
 * ** so equation will have solutions in Q1 and Q2 **


 * In Degrees: **

math . \qquad \theta = 30^\circ \; \text{ and } \; \theta = 180^\circ - 30^\circ = 150^\circ math


 * but these are only the solutions within the first cycle [0º, 360º]
 * we get more solutions by repeatedly adding 360º to each of those answers:
 * because the __**period**__ of sin( q ) is 360º

... ... ... q = {..., –330º, 30º, 390º, 750º, ...}
 * Q1: ** .... 30º, 360º + 30º, 720º + 30º, etc {also negative angles} or

... ... ... and

... ... ... q = {..., –210º, 150º, 510º, 870º, ...}
 * Q2: ** .... 150º, 360º + 150º, 720º + 150º, etc {also negative angles} or


 * In Radians: **


 * We can get a similar list in radians by repeatedly adding 2 p to each of the two initial 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

The solutions in example 1 (above) 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


 * 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.


 * When the infinite list of solutions is written in terms of a parameter, n, it is called the ** general solution **.

Solutions within a Specified Domain

 * Often we will be asked for all the solutions within a domain.
 * If we have a general solution, we can simply substitute appropriate values of n to get the list.


 * IMPORTANT NOTE: **
 * If the domain is specified in degrees, the answer is expected in degrees.
 * If the domain is specified in radians, the answer should be in radians.
 * If in doubt, write the answer in radians if you can use exact values, otherwise use degrees.

** Example 1 (cont) **
... ... Find all the solutions for sin( q ) = ½ in the domain [0, 4 p ] given that the general solution is: math . \\ . \qquad \theta = \left\{ 2n\pi + \dfrac{\pi}{6},\, 2n\pi + \dfrac{5\pi}{6},\, n \in Z \right\} math


 * ** substituting n = 0 and 1 gives: **

math . \qquad \theta = \left\{ \dfrac{\pi}{6},\, \dfrac{5\pi}{6},\, \dfrac{13\pi}{6},\, \dfrac{17\pi}{6} \right\} math

Back to General Solutions
For any defined values of a, the general solutions will be:



** Example 2 **
... ... Find the general solution of the equation: math . \qquad \sqrt{8} \cos \big( 2x \big) + 2 = 0 math


 * Solution:**

math \\.\\ . \qquad \sqrt{8} \cos \big( 2x \big) + 2 = 0 \\. \\ . \qquad \sqrt{8} \cos \big( 2x \big) =-2 math . math \\ . \qquad \cos \big( 2x \big) = -\dfrac{2}{\sqrt{8}} = -\dfrac{2}{2\sqrt{2}} \\. \\ . \qquad \cos \big( 2x \big) = -\dfrac{1}{\sqrt{2}} math
 * ** First rearrange to make cos(2x) the subject: **


 * ** Now state the general solution (notice the 2x ): **

math . \qquad 2x=2n\pi \pm \cos^{-1} \left( -\dfrac{1}{\sqrt{2}} \right),\qquad n \in Z math . math . \qquad \textit{Using exact values } \cos^{-1} \left( -\dfrac{1}{\sqrt{2}} \right)=\dfrac{3\pi}{4},\,\dfrac{5\pi}{4},\, \text{etc} \;\{ \textit{but we only need the first one} \} math . math \\ . \qquad 2x=2n\pi \pm \dfrac{3\pi}{4} \quad \{ \textit{divide entire equation by 2} \} \\. \\ . \qquad x=n\pi \pm \dfrac{3\pi}{8} \qquad n \in Z math


 * (b) ** .. ** Now state the solutions in the domain [0, 2 p ] **


 * ** Substitute various values of n into the general solution. **

math \\ . \qquad n= 0 \quad \Rightarrow \quad 0 \pm \dfrac{3\pi}{8} \quad \Rightarrow \quad -\dfrac{3\pi}{8}, \;\; \dfrac{3\pi}{8} \\. \\ . \qquad n= 1 \quad \Rightarrow \quad \pi \pm \dfrac{3\pi}{8} \quad \Rightarrow \quad \dfrac{5\pi}{8}, \;\; \dfrac{11\pi}{8} \\. \\ . \qquad n= 2 \quad \Rightarrow \quad 2\pi \pm \dfrac{3\pi}{8} \quad \Rightarrow \quad \dfrac{13\pi}{8}, \;\; \dfrac{19\pi}{8} math


 * ** Eliminate the end values because they are outside the required domain **

Solution is: math . \qquad x = \left\{ \dfrac{3\pi}{8}, \;\; \dfrac{5\pi}{8}, \;\; \dfrac{11\pi}{8} \;\; \dfrac{13\pi}{8} \right\} math

Trig Equations on the Classpad

More Examples Go to top of page flat

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