Trigonometric Equation III

Problem 206: Solve the equation $\cos \theta = \tan \theta$ .

Solution: We will use some trigonometric properties to solve this equation.

(1) $\begin{align*} \cos \theta & = \tan \theta \\ \cos \theta & = \frac{\sin \theta}{\cos \theta} \\ \cos \theta \cdot \cos \theta & = \sin \theta \\ \underbrace{\cos^2 \theta}_{\cos^2 \theta + \sin^2 \theta = 1} - \sin \theta & = 0 \\ 1 - \sin^2 \theta - \sin \theta & = 0 \\ \sin^2 \theta + \sin \theta - 1 & = 0.\end{align*}$

Notice that this is a quadratic equation. Let $x = \sin \theta$ so that

$\begin{equation*} \sin^2 \theta + \sin \theta -1 = 0 \quad \Rightarrow \quad x^2 + x -1 = 0.\end{equation*}$

Using the quadratic equation we have

(2) $\begin{equation*} x = \frac{-1 \pm \sqrt{1-4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}. \end{equation*}$

Thus $x = 0.618$ or $x = -1.618$ . However, -1.618 is not an option because this value is not in the unit circle. Therefore, we will only consider the $x = 0.618$ . That is,

(3) $\begin{equation*} \sin \theta = x = 0.618. \end{equation*}$

There are two possible $\theta$ values that yield the y-coordinates 0.0618. Thus

(4) $\begin{align*} \sin \theta = 0.618 \quad \Rightarrow \quad \theta & = \sin^{-1} (0.618) \approx 0.666 \ \text{radians} \\ \theta & = \pi - \sin^{-1} (0.618) \approx 2.475 \ \text{radians}..\end{align*}$

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