Solving trigonometric equations

In circular functions like sine and cosine, equations can have multiple solutions within a specific interval. If all possible values are considered, these functions can have infinitely many solutions. The nature of circular functions that leads to these many possible solution is demonstrated in circular functions graphs. Understanding the multiple solutions of circular functions is essential for solving real-world problems involving periodic behavior, such as waves, oscillations, and circular motion. It also forms the foundation for more advanced topics in trigonometry and calculus.

Use this page to revise the following concepts of solving trigonometric equations:

Solving Trigonometric Equations: The Principal Value

When finding solutions to circular functions it is best practice to solve for the first positive solution and using this as a reference for all subsequent results. The first positive solution is in the first quadrant which is between \([0,\frac{\pi}{2}]\).

Step 1: When the solution involves an exact value, solve the trigonometric equation to find the positive angle in the first quadrant, which is the reference angle. For all other values, use a calculator to obtain the solution.

Step 2: Use symmetry properties of the unit circle to identify values of \(0 \leq \theta \leq 2\pi\)

Sin

Worked Example 1: Positive Solutions

Find the values of \(x\) in \(x\in[0,2\pi]\) for \(\sin\left(x\right) = \frac{1}{2}\)

  1. In first quadrant: \(\sin\left(x\right) = \frac{1}{2}\), what angle \(x\) would give a result of \(\frac{1}{2}\)?
    Refer back to the exact values table.

    2. \(\begin{align}\sin\left(x\right) &= \frac{1}{2} \\ x &= \frac{\pi}{6}\end{align}\)

  2. Sine is positive in the first and second quadrant.

    3. \(x = \frac{\pi}{6}\text{ and }x = \pi - \frac{\pi}{6}\)

    \(x = \frac{\pi}{6},\ x = \frac{5\pi}{6}\)


Worked Example 2: Negative solutions

Find the value of \(x\) for \(x\in[0,2\pi]\) for \(\sin\left(x\right) = -\frac{1}{2}\)

  1. In first quadrant: \(\sin\left(x\right) = \frac{1}{2}\)

    2. \(x = \frac{\pi}{6}\)

  2. Sine is negative in the third and fourth quadrant.

    3. \[x = \pi + \frac{\pi}{6}\text{ and }x = 2\pi - \frac{\pi}{6}\]

    \[x = \frac{7\pi}{6},\ x = \frac{11\pi}{6}\]


Worked Example 3: Solutions of \(a\sin\left(n\theta\right) = b\)

Find the value of \(x\) in \(x\in[0,2\pi]\) for \(\sin\left(2x\right) = \frac{1}{2}\)

  1. Let \(z = 2x\); the equation now becomes \(\sin(z) = \frac{1}{2}\). As \(z = 2x\) the domain in which to solve becomes \(z = 2x\), as \(x \in [0,2\pi]\), then \(z = 2x \in [0,4\pi]\).
    In other words solve \(\sin\left(z\right) = \frac{1}{2}\) for \(z\in[0,4\pi]\)
  2. In first quadrant: \(\sin\left(z\right) = \frac{1}{2}\)

    3. \(z = \frac{\pi}{6}\)

  3. Sine is positive in the first and second quadrant.

    4. \(z = \frac{\pi}{6}\) and \(z = \pi - \frac{\pi}{6}\) and after \(2\pi\) units

    \(z = 2\pi + \frac{\pi}{6}\) and \(z = 3\pi - \frac{\pi}{6}\)

    Therefore, we have solutions at \[z = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}\]

  4. As \(z = 2x\)

    5. \(\begin{align}2x &= \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6} \\x &= \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}\text{ for }x\in[0,2\pi]\end{align}\)

Cos

Worked Example 1: Positive Solutions

Find the value of \(x\) in \(x\in[0,2\pi]\) for \(\cos\left(x\right) = \frac{\sqrt{2}}{2}\)

  1. In first quadrant: \(\cos\left(x\right) = \frac{\sqrt{2}}{2}\)

    2. \(x = \frac{\pi}{4}\)

  2. Cosine is positive in the first and fourth quadrant.

    3. \(x = \frac{\pi}{4}\) and \(x = 2\pi - \frac{\pi}{4}\)

    \(x = \dfrac{\pi}{4},\ x= \dfrac{7\pi}{4}\)


Worked Example 2: Negative solutions

Find the value of \(x\) in \(x\in[0,2\pi]\) for \(\cos\left(x\right) = -\frac{\sqrt{2}}{2}\)

  1. In first quadrant: \(\cos\left(x\right) = \frac{\sqrt{2}}{2}\)

    2. \(x = \frac{\pi}{4}\)

  2. Cosine is negative in the second and third quadrant.

    3. \(x = \pi - \frac{\pi}{4}\) and \(x = \pi + \frac{\pi}{4}\)

    \(x = \dfrac{3\pi}{4}, x = \dfrac{5\pi}{4}\)


Worked Example 3: Solutions of \(a\cos(n\theta) = b\)

Find the value of \(x\) in \(x\in[0,2\pi]\) for \(\cos\left(\frac{1}{2}x\right) = \frac{\sqrt{2}}{2}\)

  1. Let \(z = \frac{1}{2}x\); the equation now becomes solve for \(x\in[0,2\pi]\) for \(\cos\left(z\right) = \frac{1}{2}\).
    As \(z =\frac{1}{2}x\) the domain in which to solve becomes \(z = \frac{1}{2}x\), as \(x\in [0,2\pi]\), then \(z = \frac{1}{2},\ x\in[0,\pi]\).
    In other words solve \(\cos\left(z\right) = \frac{\sqrt{2}}{2}\) for \(z\in[0,\pi]\)
  2. In first quadrant: \(\cos\left(z\right) = \frac{\sqrt{2}}{2}\)

    3. \(z = \frac{\pi}{4}\)

  3. Cosine is positive in the first and fourth quadrant but as \(z\in[0,\pi]\)

    4. \(z = \frac{\pi}{4}\)

  4. As \(z = \frac{1}{2}x\)

    5. \(\frac{1}{2}x = \frac{\pi}{4}\)

    \(x = \dfrac{\pi}{2}\) for \(x\in[0,2\pi]\)

Tan

Worked Example 1: Positive Solutions

Find the value of \(x\) in \(x\in[0,2\pi]\) for \(\tan\left(x\right) = \sqrt{3}\)

  1. In first quadrant: \(\tan\left(x\right) = \sqrt{3}\)

    2. \(x = \frac{\pi}{3}\)

  2. Tan is positive in the first and third quadrant.

    3. \(x = \frac{\pi}{3}\) and \(x = \pi + \frac{\pi}{3}\)

    \(x = \frac{\pi}{3}, x = \frac{4\pi}{3}\)


Worked Example 2: Negative solutions

Find the value of \(x\) in \(x\in[0,2\pi]\) for \(\tan\left(x\right) = -\sqrt{3}\)

  1. In first quadrant: \(\tan\left(x\right) = \sqrt{3}\)

    2. \(x = \frac{\pi}{3}\)

  2. Tan is negative in the second and fourth quadrant.

    3. \(x = \pi - \frac{\pi}{3}\) and \(x = 2\pi - \frac{\pi}{3}\)

    \(x = \frac{2\pi}{3}, x = \frac{5\pi}{3}\)


Worked Example 3: Solutions of \(a\tan(n\theta) = b\)

Find the value of \(x\) in \(x\in[0,2\pi]\) for \(\tan\left(3x\right) = \sqrt{3}\)

  1. Let \(z = 3x\); the equation now becomes solve for \(x\in[0,2\pi] \text{of} \tan\left(z\right) = \sqrt{3}\). As \(z = 3x\) the domain in which to solve becomes \(z = 3x\ \text{as}\ x\in[0,2\pi], \text{then} z = 3 x\in[0,6\pi]\). In other words solve \(z\in[0,6\pi] \text{of} \tan\left(z\right) = \sqrt{3}\)
  2. In first quadrant: \(\tan\left(z\right) = \sqrt{3}\)

    3. \(z = \frac{\pi}{3}\)

  3. Tan is positive in the first and third quadrant and for \(z\in[0,6\pi]\)

    4. \(\begin{align} z &= \frac{\pi}{3}, \pi + \frac{\pi}{3}, 2\pi + \frac{\pi}{3}, 3\pi+ \frac{\pi}{3}, 4\pi+ \frac{\pi}{3}, 5\pi+ \frac{\pi}{3}\\ &= \frac{\pi}{3}, \frac{4\pi}{3}, \frac{7\pi}{3}, \frac{10\pi}{3}, \frac{13\pi}{3}, \frac{16\pi}{3}\end{align}\)

  4. As \(z = 3x\)

    5. \(\begin{align} 3x &= \frac{\pi}{3}, \frac{4\pi}{3}, \frac{7\pi}{3}, \frac{10\pi}{3}, \frac{13\pi}{3}, \frac{16\pi}{3}\\ x &=\frac{\pi}{9}, \frac{4\pi}{9}, \frac{7\pi}{9}, \frac{10\pi}{9}, \frac{13\pi}{9}, \frac{16\pi}{9},\text{ for }x\in[0,2\pi]\end{align}\)

The General Solution

Since there can be infinitely many solutions across the entire range where the function is defined, it is useful to be able to describe a general solution. The general solution for each trigonometric function is given by the relationship that connects the first solution to every other angle that would give the same value.

Sin

Values for sine repeat in the first and second quadrant when positive or the third and fourth quadrant when negative.

Therefore \(a\in[-1,1]\), the general solution for \(\sin\left(\theta\right) = a\), is:

\(\theta = 2\pi{n} + \sin^{-1}(a)\) and \(\theta = \left(2n + 1\right)\pi - \sin^{-1}\left(a\right)\) where \(n\in \mathbb{Z}\)

The first equation will give the first positive solution of sine (in Quadrant I) and repeats every \(2\pi\) (a full revolution). The second equation \(\theta = \left(2n + 1\right)\pi - \sin^{-1}(a)\) provides the second positive solution of sine (in Quadrant II) and appears every odd multiple of \(\pi\).

Cos

Values for cosine repeat in the first and fourth quadrants when positive or the second and third quadrant when negative.

For \(a\in[-1,1]\), the general solution for \(\cos\left(\theta\right) = a\) is:

\(\theta = 2n\pi \pm \cos^{-1}\left(a\right)\), where \(n \in \mathbb{Z}\)

Tan

Values for tan repeat in the first and third when positive or the second and fourth quadrant when negative.

For \(a\in{R}\) the general solution for \(\tan\left(\theta\right) = a\) is;

\(\theta = n\pi + \tan^{-1}\left(a\right)\), where \(n\in{Z}\)

​

Solving Trigonometric Equations: Specified Domain

To solve over a specified domain, the general solution or the principal values techniques can both be applied to obtain the same result.

Examine this in the following worked examples.

Worked Example

Find the value of \(x\) in \(x\in[\pi,3\pi]\) for \(\cos\left(2x\right) = \frac{1}{2}\)

Using the Principal Value method:

  1. Let \(z = 2x\); the equation now becomes solve for \(x\in[\pi,3\pi] \text{ of } \sin\left(z\right) = \frac{1}{2}\).
    As \(z = 2x\) the domain in which to solve becomes \(z = 2x, \text{ as } x\in[\pi,3\pi], \text{ then } z = 2x\in[2\pi,6\pi]\).
    In other words solve \(z\in[2\pi,6\pi] \text{ of } \sin\left(z\right) = \frac{1}{2}\)
  2. In first quadrant: \(\sin\left(z\right) = \frac{1}{2}\)

    3. \(z = \frac{\pi}{6}\)

  3. Sine is positive in the first and second quadrant.

    4. \(z = \frac{\pi}{6}\) and \(z = \pi - \frac{\pi}{6}\) and after \(2\pi\text{ units}\)

    \(z = 2\pi + \frac{\pi}{6}\) and \(z = 3\pi - \frac{\pi}{6}\)

    Solutions for \(z\in[2\pi,6\pi]\)  are therefore

    \(z = \frac{13\pi}{6}, \frac{17\pi}{6}, \frac{25\pi}{6}, \frac{29\pi}{6}\)

  4. As \(z = 2x\)

    5. \(\begin{align}2&x = \frac{13\pi}{6}, \frac{17\pi}{6}, \frac{25\pi}{6}, \frac{29\pi}{6} \\ &x = \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{25\pi}{12}, \frac{29\pi}{12} \text{ for } x\in[\pi,3\pi]\end{align}\)

Using the general solution:

  1. Set up general solution equation

    2. \(2x = 2n\pi + \sin^{-1}\left(\frac{1}{2}\right)\) and \(2x = \left(2n + 1\right)\pi - \sin^{-1}\left(\frac{1}{2}\right)\) where \(n\in{Z}\)

  2. Find value of \(\sin^{-1}(a)\) and solve for \(x\)

    3. \(2x = 2n\pi +\frac{\pi}{6}\) and \(2x = \left(2n + 1\right)\pi - \frac{\pi}{6}\)

    \(2x = \frac{12n\pi+\pi}{6}\) and \(2x = \frac{6(2n+1)\pi - \pi}{6}\)

    \(2x = \frac{12n\pi + \pi}{6}\) and \(2x = \frac{12n\pi + 5\pi}{6}\)

    \(x = \frac{12n\pi + \pi}{12}\) and \(x = \frac{12n\pi + 5\pi}{12}\)

  3. Substitute values for \(n\) where \(n\in{Z}\), to find solutions for the required domain.

    4. Between \([\pi,3\pi]\) we will need \(n = 0, n = 1, n = 2, n = 3\)

    \(\begin{align}&x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{25\pi}{12}, \frac{19\pi}{12}  \\ &x = \frac{13\pi}{12}, \frac{17\pi}{12},\frac{125\pi}{12},\frac{29\pi}{12} \text{ for } x\in[\pi,3\pi]\end{align}\)

Hence the same result is achieved. Either approach can help to solve for specified domains.