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

  • Solve $\sin(\theta) = \frac{1}{2}$

Graphically, this is the set of points where $y = \sin(\theta)$ intersects with $y = \dfrac{1}{2}$.

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

In Degrees: $\theta = 30^\circ \; \text{ and } \; \theta = 180^\circ - 30^\circ = 150^\circ$

but these are only the solutions within the first cycle $[0^\circ, 360^\circ ]$
we get more solutions by repeatedly adding 360 to each of those answers:
because the period of $\sin(\theta) \text{ is } 360^\circ$

Q1: …. 30, 360 + 30, 720 + 30, etc {also negative angles} or … … $\theta = \{..., –330^\circ, 30^\circ, 390^\circ, 750^\circ, ...\}$

… … … and

Q2: …. 150, 360 + 150, 720 + 150, etc {also negative angles} or … … $\theta = \{..., –210^\circ, 150^\circ, 510^\circ, 870^\circ, ... \}$

In Radians:

We can get a similar list in radians by repeatedly adding $2\pi$ to each of the two initial solutions.

Q1: …. $\left\{ \dots,\, -2\pi+\dfrac{\pi}{6},\, \dfrac{\pi}{6},\, 2\pi+\dfrac{\pi}{6},\, 4\pi+\dfrac{\pi}{6},\, \dots \right\}$ … … $= \left\{ \dots,\, -\dfrac{11\pi}{6},\, \dfrac{\pi}{6},\,\dfrac{13\pi}{6},\, \dfrac{25\pi}{6},\, \dots \right\}$

… … … and

Q2: …. $\left\{ \dots,\, -2\pi+\dfrac{5\pi}{6},\, \dfrac{5\pi}{6},\, 2\pi+\dfrac{5\pi}{6},\, 4\pi+\dfrac{5\pi}{6},\, \dots \right\}$ … … $= \left\{ \dots,\, -\dfrac{7\pi}{6},\, \dfrac{5\pi}{6},\,\dfrac{17\pi}{6},\, \dfrac{29\pi}{6},\, \dots \right\}$

The solutions in example 1 (above) can be written as:

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

… … Or

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


  • 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.

General Solutions


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.


  • 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 $\sin(\theta) = \dfrac{1}{2} \text{ in } \theta \in [0, 4\pi]$ given that the general solution is:

  • $\theta = \left\{ 2n\pi + \dfrac{\pi}{6},\, 2n\pi + \dfrac{5\pi}{6},\, n \in Z \right\}$

substituting n = 0 and 1 gives:

  • $\theta = \left\{ \dfrac{\pi}{6},\, \dfrac{5\pi}{6},\, \dfrac{13\pi}{6},\, \dfrac{17\pi}{6} \right\}$

Back to General Solutions

For any defined values of a, the general solutions will be:


Example 2

(a) Find the general solution of the equation: $\sqrt{8} \cos \big( 2x \big) + 2 = 0$
(b) and hence state the solutions in the domain $x \in [0, 2\pi]$


First rearrange to make cos(2x) the subject:

$\sqrt{8} \cos \big( 2x \big) + 2 = 0$

$\sqrt{8} \cos \big( 2x \big) =-2$

$\cos \big( 2x \big) = -\dfrac{2}{\sqrt{8}} = -\dfrac{2}{2\sqrt{2}}$

$\cos \big( 2x \big) = -\dfrac{1}{\sqrt{2}}$

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

$2x=2n\pi \pm \cos^{-1} \left( -\dfrac{1}{\sqrt{2}} \right),\qquad n \in Z$

Using exact values: $\quad \cos^{-1} \left( -\dfrac{1}{\sqrt{2}} \right)=\dfrac{3\pi}{4},\,\dfrac{5\pi}{4} \qquad \textit{but we only need the first one}$

$2x=2n\pi \pm \dfrac{3\pi}{4} \quad \{ \textit{divide entire equation by 2} \}$

$x=n\pi \pm \dfrac{3\pi}{8} \qquad n \in Z$

(b) .. Now state the solutions in the domain $x \in [0, 2\pi]$

Substitute various values of n into the general solution.

$n= 0 \quad \Rightarrow \quad 0 \pm \dfrac{3\pi}{8} \quad \Rightarrow \quad -\dfrac{3\pi}{8}, \;\; \dfrac{3\pi}{8}$

$n= 1 \quad \Rightarrow \quad \pi \pm \dfrac{3\pi}{8} \quad \Rightarrow \quad \dfrac{5\pi}{8}, \;\; \dfrac{11\pi}{8}$

$n= 2 \quad \Rightarrow \quad 2\pi \pm \dfrac{3\pi}{8} \quad \Rightarrow \quad \dfrac{13\pi}{8}, \;\; \dfrac{19\pi}{8}$

Eliminate the end values because they are outside the required domain

Solution is:

$x = \left\{ \dfrac{3\pi}{8}, \;\; \dfrac{5\pi}{8}, \;\; \dfrac{11\pi}{8} \;\; \dfrac{13\pi}{8} \right\}$

Trig Equations on the Classpad

More Examples

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