## T7 Trigonometric equations

If you know the value of a trigonometric function, how do I find all the possible angles that satisfy this expression? The calculator may only give you one answer to an inverse trig question between 0 and 90 degrees (say InvCos = 40°). The unit circle can help you visualise the many different solutions for finding your angle α.

This module describes how to solve trigonometric equations over a prescribed domain. An example of such an equation is: Solve \begin{align*} \cos\theta & =0.5 \end{align*} if $$-\pi\leq\theta\leq3\pi$$.

Solving these types of problem is easier if you can identify the values of trigonometric functions on a unit circle as shown in the example below. You also need to understand angles in degrees and radians.

### Example 1

Given that $$\sin\theta=0.3,$$ find all values of $$\theta$$ in the domain $$\left\{ \theta:0^{\circ}\leq\theta\leq360^{\circ}\right\} .$$

Solution:

Draw a picture showing the possible positive values for $$\theta$$ as shown below:

There are at least two solutions shown in red and blue. We have $$\theta_{r}$$ (in red) which is given by \begin{align*} \theta_{r} & =\sin^{-1}\left(0.3\right)\\ & =17.46^{\circ}.\text{ (from calculator)} \end{align*} The other angle $$\theta_{b}$$ (in blue) may be found by symmetry: \begin{align*} \theta_{b} & =180^{\circ}-\theta_{r}\\ & =180^{\circ}-17.46^{\circ}\\ & =162.54^{\circ}. \end{align*} Hence the solutions are $$\theta=17.46^{\circ},\,162.54^{\circ}.$$

### Example 2

Given that $$\sin\theta=0.3,$$ find all values of $$\theta$$ such that $$0^{\circ}\leq\theta\leq500^{\circ}.$$

Solution:

Note that this is the same problem as in Example 1 except that the domain in which we are looking for solutions has been extended to $$500^{\circ}.$$

Draw a picture showing the possible positive values for $$\theta$$ as shown below:

From Example 1 we know $$\theta=17.46^{\circ}$$or $$162.54^{\circ}.$$ If we add $$360^{\circ}$$ to each of these, they will still satisfy $$\sin\theta=0.3.$$ The question is, are they in the domain of interest?

Adding $$360^{\circ}$$ to the solution $$\theta=17.46^{\circ}$$ found in Example 1 gives \begin{align*} \theta & =17.46^{\circ}+360^{\circ}\\ & =377.46^{\circ} \end{align*} which is less than $$500^{\circ}$$ and so a solution to Example 2. Adding $$360^{\circ}$$ to the other solution to Example $$1$$ gives \begin{align*} \theta & =162.54^{\circ}+360^{\circ}\\ & =522.54^{\circ} \end{align*} which is bigger than $$500^{\circ}$$ and so not a solution to Example 2.

Hence the solutions are $$\theta=17.46^{\circ},\,162.54^{\circ}$$ and $$377.46^{\circ}.$$

### Example 3

Solve $$\cos\alpha=0.5$$ over the domain $$\left\{ \alpha:0\leq\alpha\leq2\pi\right\} .$$

Solution:

Draw a picture showing the possible positive values for $$\alpha$$ as shown below:

Referring to the figure, \begin{align*} \alpha_{1} & =\cos^{-1}\left(0.5\right)\\ & =\frac{\pi}{3}\ (\text{from calculator)} \end{align*} and from symmetry of the unit circle, \begin{align*} \alpha_{2} & =2\pi-\alpha_{1}\\ & =2\pi-\frac{\pi}{3}\\ & =\frac{5\pi}{3}. \end{align*} Hence the solution is $$\alpha=\pi/3$$ and $$5\pi/3.$$

### Example 4

Solve $$\cos\alpha=0.5$$ over the domain $$\left\{ \alpha:-2\pi\leq\alpha\leq2\pi\right\} .$$

Solution:

Note this is the same problem as in Example 3 but the domain has been extended.

Draw a picture showing the possible positive and negative values for $$\alpha$$ as shown below:

The two positive angles (shown in red) are the same as those in Example 3, namely \begin{align*} \alpha_{1} & =\frac{\pi}{3}\\ \alpha_{2} & =\frac{5\pi}{3}. \end{align*} The negative angles (shown in blue) are $$\alpha_{3}$$ and $$\alpha_{4}.$$ Using the results from Example 3 and symmetry we have \begin{align*} \alpha_{3} & =-\alpha_{1}\\ & =-\frac{\pi}{3} \end{align*} and \begin{align*} \alpha_{4} & =-\left(2\pi-\alpha_{1}\right)\\ & =\alpha_{1}-2\pi\\ & =\frac{\pi}{3}-2\pi\\ & =-\frac{5\pi}{3}. \end{align*} Hence the solution is $$\alpha=-\frac{5\pi}{3},\,-\frac{\pi}{3},\,\frac{\pi}{3},\,\frac{5\pi}{3}.$$

### Example 5

Solve $$\tan\left(2x\right)=-3$$ over the domain $$\left\{ x:-90^{\circ}\leq x\leq180^{\circ}\right\} .$$

Solution:

Since the variable is $$2x,$$ we have to redefine the domain to $$\left\{ 2x:-180^{\circ}\leq2x\leq360^{\circ}\right\} .$$

Draw a picture showing the possible positive and negative values for $$2x$$ as shown below:

There are two positive angles $$\alpha_{1},\,\alpha_{2}$$ (shown in red) and two negative angles $$\alpha_{3,}\,\alpha_{4}$$ (shown in blue) that may lie within $$-180^{\circ}\leq2x\leq360^{\circ}.$$

We have \begin{align*} \tan\left(2x\right) & =-3\\ 2x & =\tan^{-1}\left(-3\right)\\ & =-71.57^{\circ}\ \text{(from calculator).} \end{align*} Referring to the figure, we see that \begin{align*} \alpha_{3} & =-71.57^{\circ} \end{align*} and \begin{align*} \alpha_{4} & =\alpha_{3}-180^{\circ}\\ & =-71.57^{\circ}-180^{\circ}\\ & =-251.57^{\circ}. \end{align*} Note that this is outside of the domain for $$2x$$ and so is rejected. That is we discard $$\alpha_{4}.$$

Using the symmetry of the unit circle, we have \begin{align*} \alpha_{1} & =360^{\circ}+\alpha_{4}\\ & =360^{\circ}-251.57^{\circ}\\ & =108.43^{\circ} \end{align*} and using the figure, \begin{align*} \alpha_{2} & =\alpha_{1}+180^{\circ}\\ & =108.43^{\circ}+180^{\circ}\\ & =288.43^{\circ}. \end{align*} Hence \begin{align*} 2x & =\alpha_{3},\ \alpha_{1},\ \alpha_{2}\\ & =-71.57^{\circ},\ 108.43^{\circ},\ 288.43^{\circ} \end{align*} and the solution is \begin{align*} x & =-35.79^{\circ},\ 54.22^{\circ},\ 144.22^{\circ}. \end{align*}

### Exercises

Solve the following equations:

$$\text{a) If \sin\phi=0.25 find \phi for 0^{\circ}\leq\phi\leq180^{\circ} .}$$

$$\text{b) If \tan\phi=0.8 find }\phi$$ for $$0^{\circ}\leq\phi\leq360^{\circ}$$.

$$\text{c) If \cos\phi=0.4 find \phi\ \text{for 0^{\circ}\leq\phi\leq360^{\circ}. } }$$

$$\text{d) If \cos\phi=-0.4\ \text{find \phi\text{ for -180^{\circ}\leq\phi\leq360^{\circ}. } } }$$

$$\text{e) If \tan\phi=-1.5\ \text{find \phi\ \text{for -180^{\circ}\leq\phi\leq360^{\circ}. } } }$$

$$\text{f) If \cos\phi=-0.3\ \text{find \phi for 0^{\circ}\leq \phi\leq360^{\circ}. } }$$

$$\text{a) 14.5^{\circ},\,165.5^{\circ} \qquad\text{b) 38.7^{\circ},\,218.7^{\circ}\qquad\text{c) 66.4^{\circ},\,293.6^{\circ} } } }$$
$$\text{d) -113.6^{\circ},\,113.6^{\circ},\,246.4^{\circ}\qquad\text{e) -56.3^{\circ},\,123.7^{\circ},\,303.7^{\circ}\qquad\text{f) 107.5^{\circ},\,252.5^{\circ}. } } }$$