Section 6.5: Solving Trigonometric Equations Using Identities

We see immediately in the equation $\sec^{2}(\theta) = \tan(\theta) + 3$ that there are two different circular functions present, so we look for an identity to express both sides in terms of the same function.

We use the Pythagorean Identity $\sec^{2}(\theta) = 1 + \tan^{2}(\theta)$ to exchange $\sec^{2}(\theta)$ for tangents. What results is a `quadratic in disguise', which we can solve using a quick substitution.

\[ \begin{array}{rclr} \sec^{2}(\theta) & = & \tan(\theta) + 3 & \\ 1 + \tan^{2}(\theta) & = & \tan(\theta) + 3& \text{(Since $\sec^{2}(\theta) = 1 + \tan^{2}(\theta)$.)} \\ \tan^{2}(\theta) - \tan(\theta) -2 & = & 0 & \\ u^2 - u - 2 & = & 0 & \text{Let $u = \tan(\theta)$.} \\ (u + 1)(u - 2) & = & 0 & \\ \end{array} \]

This gives $u = -1$ or $u = 2$. Since $u = \tan(\theta)$, we have $\tan(\theta) = -1$ or $\tan(\theta) = 2$.

From $\tan(\theta) = -1$, we get $\theta = -\frac{\pi}{4} + \pi k$ for integers $k$. To solve $\tan(\theta) = 2$, we employ the arctangent function and get $\theta = \arctan(2) + \pi k$ for integers $k$.

From the first set of solutions, we get $\theta = \frac{3\pi}{4}$ and $\theta = \frac{7\pi}{4}$ as our answers which lie in $[0,2\pi)$.

We get $\theta=\arctan(2) \approx 1.107$ and $\theta = \pi + \arctan(2) \approx 4.249$ as answers from our second set of solutions which lie in $[0,2\pi)$.

The good news is that in the equation $\cos(2t) = 3\cos(t) - 2$, we have the same circular function, cosine, throughout. The bad news is that we have different arguments, $2t$ and $t$.

Using the double angle identity $\cos(2t) = 2\cos^{2}(t) - 1$ results in another quadratic in disguise:'

\[ \begin{array}{rclr} \cos(2t) & = & 3\cos(t) - 2 & \\ 2\cos^{2}(t) -1 & = & 3\cos(t) -2 & \text{(Since $\cos(2t) = 2\cos^{2}(t) -1$.)} \\ 2\cos^{2}(t) - 3\cos(t) +1 & = & 0 & \\ 2 u^2 - 3 u + 1 & = & 0 & \text{Let $u = \cos(t)$.}\\ (2u - 1)(u - 1) & = & 0 & \\ \end{array} \]

We get $u = \frac{1}{2}$ or $u = 1$, so $\cos(t) = \frac{1}{2}$ or $\cos(t) = 1$. Solving $\cos(t) = \frac{1}{2}$, we get $t = \frac{\pi}{3} + 2\pi k$ or $t = \frac{5\pi}{3} + 2\pi k$ for integers $k$. From $\cos(t) = 1$, we get $t = 2\pi k$ for integers $k$. The answers which lie in $[0,2\pi)$ are $t =0$, $\frac{\pi}{3}$, and $\frac{5\pi}{3}$.

To solve $\cos(3t) = 2- \cos(t)$, we take a cue from the previous problem and look for an identity to rewrite $\cos(3t)$ in terms of $\cos(t)$.

We can start with a sum identity and work for there. Remember, the goal is to end up with only $\cos(t)$ in our expression:

$$\begin{array}{lr} \cos(3t) &\\ &\\ =\cos(2t + t) &\\ &\\ =\cos(2t)\cos(t)-\sin(2t)\sin(t) &\text{Cosine sum identity}\\ &\\ =(2\cos^2(t)-1)\cos(t)-2\sin(t)\cos(t)\sin(t) &\text{Double angle identities}\\ &\\ =2\cos^3(t)-\cos(t)-2\sin^2(t)\cos(t) &\\ &\\ =2\cos^3(t)-\cos(t)-2(1-\cos^2(t))\cos(t) &\text{Pythagorean Identity}\\ &\\ =2\cos^3(t)-\cos(t)-2\cos(t)+2\cos^3(t) &\\ &\\ =4\cos^3(t)-3\cos(t)&\\ \end{array}$$

This transforms the equation into a polynomial in terms of $\cos(t)$. \[ \begin{array}{rclr} \cos(3t) & = &2- \cos(t) & \\ 4\cos^{3}(t) - 3\cos(t) & = & 2- \cos(t) & \\ 2\cos^{3}(t) - 2\cos(t) -2 & = & 0 & \\ 4 u^3 - 2 u -2 & = & 0 & \text{Let $u = \cos(t)$.} \\ \end{array} \]

Using what we know about polynomials, we factor $4u^3-2u-2$ as $(u-1)\left(4u^2+4u+2\right)$ and set each factor equal to $0$.

We get either $u-1 = 0$ or $4u^2+2u+2=0$, and since the discriminant of the latter is negative, the only real solution to $4u^3-2u-2=0$ is $u = 1$.

Since $u = \cos(t)$, we get $\cos(t) = 1$, so $t = 2\pi k$ for integers $k$. The only solution which lies in $[0,2\pi)$ is $t = 0$.

While we could approach solving the equation $\cos(3x) = \cos(5x)$ in the same manner as we did the previous two problems, we choose instead to showcase the utility of the Sum to Product Identities.

From $\cos(3x) = \cos(5x)$, we get $\cos(5x) - \cos(3x) = 0$, and it is the presence of $0$ on the right hand side that indicates a switch to a product would be a good move.

Using the Sum-to-Product identities, we rewrite $\cos(5x) - \cos(3x)$ as $- 2 \sin\left( \frac{5x + 3x}{2}\right)\sin\left( \frac{5x - 3x}{2}\right) = -2 \sin(4x)\sin(x)$. Hence, our original equation $\cos(3x) = \cos(5x)$ is equivalent to $-2 \sin(4x) \sin(x) = 0$.

From $-2 \sin(4x) \sin(x) = 0$, thanks to the Zero Product Property, we get either $\sin(4x) = 0$ or $\sin(x)$ = 0. Solving $\sin(4x) = 0$ gives $x = \frac{\pi}{4} k$ for integers $k$, and the solution to $\sin(x) = 0$ is $x = \pi k$ for integers $k$.

The second set of solutions is contained in the first set of solutions,\footnote{As always, when in doubt, write it out!} so our final solution to $\cos(5x) = \cos(3x)$ is $x = \frac{\pi}{4} k$ for integers $k$.

There are eight of these answers which lie in $[0,2\pi)$: $x = 0$, $\frac{\pi}{4}$, $\frac{\pi}{2}$, $\frac{3\pi}{4}$, $\pi$, $\frac{5\pi}{4}$, $\frac{3\pi}{2}$ and $\frac{7\pi}{4}$.