Solving trigonometric equations

You might notice that this is the same problem as in Example 1, but with the domain extended to \(500^{\circ}\).

In the same way, we can sketch a unit circle and locate the points for which \(\sin(\theta)=0.3\).

From Example 1, we know \(\theta=17.46^{\circ}\) or \(162.54^{\circ}\). If we add \(360^{\circ}\) to each of these, i.e. go around the unit circle a second time, they will still satisfy \(\sin\theta=0.3\) function. The question is, are they in the domain of interest?

\[\begin{align*} \theta & = 17.46^{\circ}+360^{\circ}\\
& = 377.46^{\circ}
\end{align*}\]

This is less than \(500^{\circ}\) and is therefore a solution to Example 2.

\[\begin{align*} \theta & = 162.54^{\circ}+360^{\circ}\\
& = 522.54^{\circ}
\end{align*}\]

This is greater than \(500^{\circ}\), and goes beyond the domain, so it is not a solution to Example 2.

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

Sketch a unit circle and locate the points on the unit circle for which \(\cos\alpha=0.5\)—that is, where \(x=0.5\).

There are two possible solutions, \(\alpha_{1}\) and \(\alpha_{2}\).

\[\begin{align*} \alpha_{1} & = \cos^{-1}\left(0.5\right)\\
& = \frac{\pi}{3}\quad\textrm{from calculator}
\end{align*}\]

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 solutions are \(\alpha=\dfrac{\pi}{3}\) and \(\dfrac{5\pi}{3}\).

You might notice that this is the same problem as in Example 3, but with the domain extended to \(-2\pi\).

In the same way, we can sketch a unit circle and locate the points for which \(\cos(\alpha)=0.5\).

The two positive angles (shown in red) are the same as those in Example 3: \(\alpha_{1}=\dfrac{\pi}{3}\) and \(\alpha_{2}=\dfrac{5\pi}{3}\).

The negative angles (shown in blue) are \(\alpha_{3}\) and \(\alpha_{4}\).

Using the results from Example 3 and symmetry, we have:

\[\begin{align*} \alpha_{3} & = -\alpha_{1}\\
& = -\frac{\pi}{3}\\
\alpha_{4} & = -\left(2\pi-\alpha_{1}\right)\\
& = \alpha_{1}-2\pi\\
& = \frac{\pi}{3}-2\pi\\
& = -\frac{5\pi}{3}
\end{align*}\]

Hence, the solutions are \(\alpha=-\dfrac{5\pi}{3},\,-\dfrac{\pi}{3},\,\dfrac{\pi}{3}\) and \(\dfrac{5\pi}{3}\).

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

Points for tangent functions can be a bit more difficult to locate on a unit circle at first, so what we can do is find the first angle using our calculator.

\[\begin{align*} \tan\left(2x\right) & = -3\\
2x & = \tan^{-1}\left(-3\right)\\
& = -71.57^{\circ}\quad\textrm{from calculator} \end{align*}\]

We can use this angle to draw a line that cuts through the origin \((0,0)\), then find the other angles that lie within \(-180^{\circ} \le 2x \le 360^{\circ}\).

There are two positive angles \(\alpha_{1}\) and \(\alpha_{2}\) (shown in red) and two negative angles \(\alpha_{3,}\) and \(\alpha_{4}\) (shown in blue).

We can see that \(\alpha_{3}=-71.57^{\circ}\).

Using symmetry, we can find \(\alpha_{4}\).

\[\begin{align*} \alpha_{4} & = \alpha_{3}-180^{\circ}\\
& = -71.57^{\circ}-180^{\circ}\\
& = -251.57^{\circ}
\end{align*}\]

We can see that this is outside of the domain for \(2x\), so the answer is rejected.

Using the symmetry of the unit circle again, we can find \(\alpha_{1}\) and \(\alpha_{2}\).

\[\begin{align*} \alpha_{1} & = 360^{\circ}+\alpha_{4}\\
& = 360^{\circ}-251.57^{\circ}\\
& = 108.43^{\circ}\\
\alpha_{2} & = \alpha_{1}+180^{\circ}\\
& = 108.43^{\circ}+180^{\circ}\\
& = 288.43^{\circ}
\end{align*}\]

Hence:

\[\begin{align*} 2x & = \alpha_{3},\ \alpha_{1},\ \alpha_{2}\\
& = -71.57^{\circ},\ 108.43^{\circ},\ 288.43^{\circ}
\end{align*}\]

The solutions are \(x=-35.79^{\circ},\ 54.22^{\circ}\) and \(144.22^{\circ}\).