Section 5.6: Solving Trigonometric Equations

Since $\frac{1}{2}$ is one of our special unit circle values, we start by finding solutions by inspection. That is, we look at the unit circle and recall which angles have a cosine of $\frac{1}{2}$. Those angles, in QI and QIV, are $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$.

These are not the only solutions, however. Take $\theta = \frac{\pi}{3}$. Any other coterminal angle will also have a cosine of $\frac{1}{2}$. That means $\frac{\pi}{3}+2\pi = \frac{7\pi}{3}$ and $\frac{7\pi}{3}+2\pi = \frac{13\pi}{3}$ both have cosines of $\frac{1}{2}$ and are therefore solutions. In fact, we can keep adding (or subtracting) $2\pi$ infinitely to come up with more solutions! Hence to describe all angles coterminal with a given angle, we add $2\pi k$ for integers $k = 0$, $\pm 1$, $\pm 2$, \dots. Hence, we record our final answer as $\theta = \frac{\pi}{3} + 2\pi k$ for integers $k$.

Proceeding similarly for the Quadrant IV case, we find the solution to $\cos(\theta) = \frac{1}{2}$ here is $\frac{5 \pi}{3}$, so our answer in this Quadrant is $\theta = \frac{5\pi}{3} + 2\pi k$ for integers $k$.

Again using inspection, we see that the common unit circle values of $\alpha = \frac{7\pi}{6}$ and $\frac{11\pi}{6}$ have sines of $-\frac{1}{2}$. We capture all solutions by adding multiples of $2\pi$ to create coterminal angles: $$ \alpha = \frac{7\pi}{6} + 2\pi k$$ $$\alpha = \frac{11\pi}{6} + 2\pi k$$ for integers $k$.

To solve $\sec(\theta) = 2$, we convert to cosines and get $\frac{1}{\cos(\theta)} = 2$ or $\cos(\theta) = \frac{1}{2}$. The answer is: $\theta = \frac{\pi}{3} + 2\pi k$ or $\theta = \frac{5\pi}{3} + 2\pi k$ for integers $k$.

The inspection bit is a little harder on this one, but with some work we can find that the angles $\theta = \frac{\pi}{3}$ and $\frac{4\pi}{3}$ have tangents of $\sqrt{3}$. In Quadrant I, we get the solutions: $\theta = \frac{\pi}{3} + 2\pi k$ for integers $k$, and for Quadrant III, we get $\theta = \frac{4\pi}{3} + 2\pi k$ for integers $k$. While these descriptions of the solutions are correct, they can be combined into one list as $\theta = \frac{\pi}{3} + \pi k$ for integers $k$.

First we must isolate the $\cos(\alpha)$: $$2\cos(\alpha) = -2$$ $$\cos(\alpha) = -1$$ Now we can proceed with inspection. We see that $\alpha = \pi$ has a cosine of -1. Our final set of solutions is $\alpha = \pi + 2\pi k$ for all integers $k$.

Here, we're being asked only for the solutions that are the initial special angles on the unit circle. That is, $$\alpha = \frac{\pi}{6}, \frac{11\pi}{6}$$ We don't need to add any copies of $2\pi$, because we've found all the solutions between 0 and $2\pi$ just by inspection!

We start in the same place, with $\alpha = \frac{\pi}{6}, \frac{11\pi}{6}$. However, we want all valid solutions all the way up to $4\pi$, so we need to add $2\pi$ to find some coterminal angles: $$\alpha = \frac{\pi}{6}+2\pi = \frac{\pi}{6}+\frac{12\pi}{6} = \frac{13\pi}{6}$$ $$\alpha = \frac{11\pi}{6}+2\pi = \frac{23\pi}{6}$$ We stop here, because if we were to add any more cycles of $2\pi$, we would create angles larger than $4\pi$. In other words, since we want our solutions to be between 0 and $4\pi$, we only need to look at two rotations of the unit circle.

Our final set of answers is $\alpha = \frac{\pi}{6}, \frac{11\pi}{6}, \frac{13\pi}{6}, \frac{23\pi}{6}$.

Here, only our first solution, $\frac{\pi}{6}$ initially falls in the relevant interval. Since we want to look at negative angles, we can create coterminal angles by subtracting $2\pi$: $$\alpha = \frac{\pi}{6} - 2\pi = -\frac{11\pi}{6}.$$

That solution doesn't work; it's to the left of $-\pi$. Let's try our other initial special value instead; that is, even though it doesn't fall in the right interval initially, when we solve by inspection we see that $\alpha = \frac{11\pi}{6}$ has a cosine of $\frac{\sqrt{3}}{2}$, so let's find a negative coterminal angle: $$\alpha = \frac{11\pi}{6}-2\pi = -\frac{\pi}{6}$$ That one works! Our final solution set is $$\alpha = \frac{\pi}{6}, -\frac{\pi}{6}.$$

This equation is at least in the form $\cos(u) = -\frac{\sqrt{3}}{2}$ if we let $u = 2\theta$. In that case, The solutions to $\cos(u) =-\frac{\sqrt{3}}{2}$ are $u = \frac{5\pi}{6} + 2\pi k$ or $u = \frac{7\pi}{6} + 2\pi k$ for integers $k$.

Since the argument of cosine here is $2\theta$, this means $2\theta = \frac{5\pi}{6} + 2\pi k$ or $2\theta = \frac{7\pi}{6} + 2\pi k$ for integers $k$. Solving for $\theta$ gives $\theta = \frac{5\pi}{12} + \pi k$ or $\theta = \frac{7\pi}{12} + \pi k$ for integers $k$.

To check these answers analytically, we substitute them into the original equation. For any integer $k$: \[ \begin{array}{rclr} \cos\left( 2\left[\frac{5\pi}{12} + \pi k\right]\right) & = & \cos\left(\frac{5\pi}{6} + 2\pi k\right) & \\ & = & \cos\left(\frac{5\pi}{6}\right) & \text{(the period of cosine is $2\pi$)} \\ & = & -\frac{\sqrt{3}}{2} & \\ \end{array}\]

Similarly, we find $\cos\left( 2\left[\frac{7\pi}{12} + \pi k\right]\right) = \cos\left(\frac{7\pi}{6} + 2\pi k\right) = \cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2}$.

To determine which of our solutions lie in $[0,2\pi)$, we substitute integer values for $k$. The solutions we keep come from the values of $k = 0$ and $k =1$ and are $\theta = \frac{5\pi}{12}$, $\frac{7\pi}{12}$, $\frac{17\pi}{12}$ and $\frac{19\pi}{12}$.

Again, we let $u = 2\theta$, but this time before we find our solutions, we consider that if $$0\leq \theta < 2\pi$$ then $$2\cdot 0 \leq 2 \cdot \theta < 2\cdot 2\pi$$ $$\Rightarrow 0 \leq u < 4\pi$$ So, if we want to solve $\cos(u) = -\frac{\sqrt{3}}{2}$, we need to solve on the interval $[0, 4\pi)$, or two rotations of the unit circle.

The first set of solutions comes from $u = \frac{5\pi}{6}$ and its coterminal pal $u = \frac{5\pi}{6}+2\pi = \frac{17\pi}{6}$. The next set come from $u = \frac{7\pi}{6}$ and $u \frac{7\pi}{6}+2\pi = \frac{19\pi}{6}$.

Now we put things back in terms of the original $\theta$: $$2\theta = \frac{5\pi}{6} \Rightarrow \theta = \frac{5\pi}{12}$$ $$2\theta = \frac{7\pi}{6} \Rightarrow \theta = \frac{7\pi}{12}$$ $$2\theta = \frac{17\pi}{6} \Rightarrow \theta = \frac{17\pi}{12}$$ $$2\theta = \frac{19\pi}{6} \Rightarrow \theta = \frac{19\pi}{12}$$

Since this equation has the form $\csc(u) = \sqrt{2}$, we rewrite this as $\sin(u) = \frac{\sqrt{2}}{2}$ and find $u = \frac{\pi}{4} + 2\pi k$ or $u = \frac{3\pi}{4} + 2\pi k$ for integers $k$.

Since the argument of cosecant here is $\left(\frac{1}{3}\theta-\pi \right)$, $\frac{1}{3}\theta-\pi = \frac{\pi}{4} + 2\pi k$ or $\frac{1}{3}\theta - \pi = \frac{3\pi}{4} + 2\pi k$.

To solve $\frac{1}{3} \theta-\pi = \frac{\pi}{4} + 2\pi k$, we first add $\pi$ to both sides to get $\frac{1}{3} \theta = \frac{\pi}{4} + 2\pi k + \pi$. A common error is to treat the `$2\pi k$' and `$\pi$' terms as `like' terms and try to combine them when they are not.

We can, however, combine the `$\pi$' and `$\frac{\pi}{4}$' terms to get $\frac{1}{3} \theta = \frac{5\pi}{4} + 2\pi k$.

We now finish by multiplying both sides by $3$ to get $\theta = 3 \left( \frac{5\pi}{4} + 2\pi k \right) = \frac{15 \pi}{4} + 6\pi k$, where $k$, as always, runs through the integers.

Solving the other equation, $\frac{1}{3} \theta-\pi = \frac{3\pi}{4} + 2\pi k$ produces $\theta = \frac{21\pi}{4} + 6 \pi k$ for integers $k$. To check the first family of answers, we substitute, combine line terms, and simplify.

\[ \begin{array}{rclr} \csc\left(\frac{1}{3} \left[ \frac{15\pi}{4} + 6 \pi k \right] - \pi \right) & = & \csc\left(\frac{5\pi}{4} + 2\pi k - \pi \right) & \\ & = & \csc\left(\frac{\pi}{4} + 2\pi k\right) & \\ & = & \csc\left(\frac{\pi}{4}\right) & \text{(the period of cosecant is $2\pi$)} \\ & = & \sqrt{2} & \\ \end{array}\] The family $\theta = \frac{21\pi}{4} + 6 \pi k$ checks similarly.

Despite having infinitely many solutions, we find that none of them lie in $[0,2\pi)$.

First we let $u = \frac{1}{3}\theta - \pi$ and find that if $$0 \leq \theta < 2\pi$$ $$\frac{1}{3}\cdot 0 - \pi \leq \frac{1}{3}\theta - \pi < \frac{1}{3}\cdot 2\pi - \pi$$ $$\Rightarrow -\pi \leq u < -\frac{\pi}{3}$$

That's a pretty small interval in which to search for our solution, but we'll check.

We know if $\csc(u) = \sqrt{2}$ then $\frac{1}{\sin(u)} = \frac{\sqrt{2}}{2}$, so our candidates for solutions are $u = \frac{\pi}{4}$ and $\frac{3\pi}{4}$. To see if these have negative coterminal angles that fall between $-\pi$ and $-\frac{\pi}{3}$, we subtract a full rotation, $2\pi$. $$\frac{\pi}{4}-2\pi = \frac{\pi}{4}-\frac{8\pi}{4} = -\frac{7\pi}{4}$$ $$\frac{3\pi}{4}-2\pi = -\frac{5\pi}{4}$$

Neither of these angles falls between $-\pi$ and $-\frac{\pi}{3}$, so our search stops here; there are no solutions in the interval.

Since $\cot(3t) = 0$ has the form $\cot(u) = 0$, we know $u = \frac{\pi}{2} + \pi k$, so, in this case, $3t = \frac{\pi}{2} + \pi k$ for integers $k$.

Solving for $t$ yields $t = \frac{\pi}{6} + \frac{\pi}{3} k$. Checking our answers, we get \[ \begin{array}{rclr} \cot\left(3\left[ \frac{\pi}{6} + \frac{\pi}{3} k\right]\right) & = & \cot\left(\frac{\pi}{2} + \pi k\right) & \\ & = & \cot\left(\frac{\pi}{2}\right) & \text{(the period of cotangent is $\pi$)} \\ & = & 0 & \\ \end{array}\]

As $k$ runs through the integers, we obtain six answers, corresponding to $k=0$ through $k=5$, which lie in $[0, 2\pi)$: $x = \frac{\pi}{6}$, $\frac{\pi}{2}$, $\frac{5\pi}{6}$, $\frac{7\pi}{6}$ , $\frac{3\pi}{2}$ and $\frac{11\pi}{6}$.

If we let $u = 3t$, then $$0\leq t < 2\pi$$ $$\Rightarrow 3\cdot 0 \leq 3t < 3\cdot 2\pi$$ $$0 \leq u < 6\pi$$

So this time, we want to solve $\cot(u) = 0$ by looking at three rotations of the unit circle. Our first set of solutions comes from when $u = \frac{\pi}{2}$: $$u = \frac{\pi}{2}$$ $$u = \frac{\pi}{2}+2\pi = \frac{5\pi}{2}$$ $$u = \frac{5\pi}{2}+2\pi = \frac{9\pi}{2}$$

For the second batch, we start with $u = \frac{3\pi}{2}$: $$u = \frac{3\pi}{2}$$ $$u = \frac{3\pi}{2}+2\pi = \frac{7\pi}{2}$$ $$u = \frac{7\pi}{2}+2\pi = \frac{11\pi}{2}$$

In each of these cases, $u = 3t$, so to finish solving, we divide each of our six answers by 3, producing a final solution of $t = \frac{\pi}{6}$, $\frac{\pi}{2}$, $\frac{5\pi}{6}$, $\frac{7\pi}{6}$ , $\frac{3\pi}{2}$ and $\frac{11\pi}{6}$.

If $\sin(\theta) = \frac{1}{3}$, then the terminal side of $\theta$, when plotted in standard position, intersects the Unit Circle at $y = \frac{1}{3}$. Geometrically, we see that this happens at two places: in Quadrant I and Quadrant II.

If we let $\alpha$ denote the acute solution to the equation, then all the solutions to this equation in Quadrant I are coterminal with $\alpha$, and $\alpha$ serves as the reference angle for all of the solutions to this equation in Quadrant II as seen below.

Since $\frac{1}{3}$ isn't the sine of any of the `common angles' we've encountered, we use the arcsine functions to express our answers. By definition, real number $t = \arcsin\left(\frac{1}{3}\right)$ $\sin(t) = \frac{1}{3}$ with $0 < t < \frac{\pi}{2}$.

Hence, $\alpha = \arcsin\left(\frac{1}{3}\right)$ radians is an acute angle with $\sin(\alpha) = \frac{1}{3}$. Since all of the Quadrant I solutions $\theta$ are all coterminal with $\alpha$, we get $\theta = \alpha + 2\pi k = \arcsin\left(\frac{1}{3}\right) + 2\pi k$ for integers $k$.

Turning our attention to Quadrant II, we get one solution to be $\pi - \alpha$. Hence, the Quadrant II solutions are $\theta = \pi - \alpha + 2\pi k = \pi - \arcsin\left(\frac{1}{3}\right) + 2\pi k$, for integers $k$.

The real number solutions to $\tan(t)=-2$ correspond to angles $\theta$ with $\tan(\theta) = -2$. Since tangent is negative only in Quadrants II and IV, we focus our efforts there.

The real number $t = \arctan(-2)$ satisfies $\tan(t)=-2$ and $-\frac{\pi}{2} < t < 0$. If we let $\beta = \arctan(-2)$ radians, then all of the Quadrant IV solutions to $\tan(\theta) = -2$ are coterminal with $\beta$.

Moreover, the solutions from Quadrant II differ by exactly $\pi$ units from the solutions in Quadrant IV (recall, the period of the tangent function is $\pi$.) Hence, all of the solutions to $\tan(\theta) = -2$ are of the form $\theta = \beta + \pi k = \arctan(-2) + \pi k$ for some integer $k$. Switching back to the variable $t$, we record our final answer to $\tan(t) = -2$ as $t = \arctan(-2) + \pi k$ for integers $k$.

To solve $\sin(2x) = 0.87$, we first note that it has the form $\sin(u) = 0.87$, which has the family of solutions $u = \arcsin(0.87) + 2\pi k$ or $u =\pi - \arcsin(0.87) + 2\pi k$ for integers $k$.

Since the argument of sine here is $2x$, we get $2x = \arcsin(0.87) + 2\pi k$ or $2x =\pi - \arcsin(0.87) + 2\pi k$ which gives $x = \frac{1}{2} \arcsin(0.87) + \pi k$ or $x =\frac{\pi}{2} - \frac{1}{2}\arcsin(0.87) + \pi k$ for integers $k$.

To determine which of these solutions lie in $[0,2\pi)$, we first need to get an idea of the value of $x=\frac{1}{2} \arcsin(0.87)$. Once again, we could use the calculator, but we adopt an analytic route here.

By definition, $0 \lt \arcsin(0.87) \lt \frac{\pi}{2}$ so that multiplying through by $\frac{1}{2}$ gives us $0 \lt \frac{1}{2} \arcsin(0.87) \lt\frac{\pi}{4}$.

Starting with the family of solutions $x = \frac{1}{2} \arcsin(0.87) + \pi k$, we use the same kind of arguments as in our solution to the similar problem above and find only the solutions corresponding to $k =0$ and $k=1$ lie in $[0,2\pi)$: $x = \frac{1}{2} \arcsin(0.87)$ and $x = \frac{1}{2} \arcsin(0.87) + \pi$.

Next, we move to the family $x =\frac{\pi}{2} - \frac{1}{2}\arcsin(0.87) + \pi k$ for integers $k$. Here, we need to get a better estimate of $\frac{\pi}{2} - \frac{1}{2} \arcsin(0.87)$. From the inequality $0 \lt \frac{1}{2}\arcsin(0.87) \lt \frac{\pi}{4}$, we first multiply through by $-1$ and then add $\frac{\pi}{2}$ to get $\frac{\pi}{2} > \frac{\pi}{2} -\frac{1}{2} \arcsin(0.87) > \frac{\pi}{4}$, or $\frac{\pi}{4} \lt \frac{\pi}{2} -\frac{1}{2} \arcsin(0.87) \lt \frac{\pi}{2}$.

Proceeding with the usual arguments, we find the only solutions which lie in $[0,2\pi)$ correspond to $k = 0$ and $k=1$, namely $x =\frac{\pi}{2} - \frac{1}{2}\arcsin(0.87)$ and $x = \frac{3\pi}{2} - \frac{1}{2}\arcsin(0.87)$.

All told, we have found four solutions to $\sin(2x) = 0.87$ in $[0,2\pi)$: $x =\frac{1}{2} \arcsin(0.87) \approx 0.528$, $x=\frac{1}{2} \arcsin(0.87) + \pi \approx 3.669$, $x =\frac{\pi}{2} - \frac{1}{2}\arcsin(0.87) \approx 1.043$ and $x = \frac{3\pi}{2} - \frac{1}{2}\arcsin(0.87) \approx 4.185$.

The first thing to notice is that we have a cosine squared and a cosine in this equation. That means we can do a quick substitution to make this look like a quadratic equation. Let $u = \cos(\theta)$; then $u^2 = \cos^2(\theta)$ and our equation becomes $$u^2 + 2u - 3 = 0.$$

To solve, factor and use the Zero Product Property: $$(u +3)(u-1)=0$$ $$\Rightarrow u+3=0 \textrm{ or } u - 1 = 0$$ $$u = -3 \textrm{ or } u = 1$$

Now, we replace $u$ with the original $\cos(\theta)$ and finish solving for $\theta$ $$\cos(\theta) = -3 \textrm{ or } \cos(\theta) = 1$$

Recall that the range of cosine is $[-1,1]$, so the equation $\cos(\theta) = -3$ is extraneous; it can't have a solution!. By inspection, we know that cosine is 1 when $\theta = 0$, so our final solution is $\theta = 0$.

Once again, we can view this equation as a quadratic if we let $u = \sec(t)$: $$u^2 = 4$$

To solve for u in this case, we can take the square root, but keep in mind, that means we end up with two solutions: $$u = 2 \textrm{ or } u = -2.$$ Subsitution our secants back in, we have $$\sec(t) = 2 \textrm{ or } \sec(t) = -2$$ and putting this in terms of cosine we find that $$\cos(t) = \frac{1}{2} \textrm{ or } \cos(t) = -\frac{1}{2}.$$

Now we solve by inspection; the common angles which have a cosine of $\frac{1}{2}$ or $-\frac{1}{2}$ are $t = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$.

One approach to solving $3\sin^{3}(\theta) = \sin^{2}(\theta)$ begins with dividing both sides by $\sin^{2}(\theta)$. Doing so, however, assumes that $\sin^{2}(\theta) \neq 0$ which means we risk losing solutions. This brings us to a nice general principle:

If you are tempted to divide by a variable to solve an equation, try factoring and the Zero Product Property instead!

Let's once again make a substitution and try to solve this like a regular polynomial equation. Let $u = \sin(\theta)$. $$3u^3 = u^2$$ $$3u^3-u^2 = 0$$ $$u^2(3u-1) = 0$$

Now we apply the Zero Product Property!

$$u^2 = 0 \textrm{ or } 3u - 1 = 0$$ $$\Rightarrow u = 0, \frac{1}{3}$$

Since $u = \sin(\theta)$, $\sin(\theta) = 0$ or $\sin(\theta) = \frac{1}{3}$. The solution to $\sin(\theta) = 0$ is $\theta = 0, \pi$, for the two solutions which lie in $[0,2\pi)$.

To solve $\sin(\theta) = \frac{1}{3}$, we use the arcsine function to get $\theta = \arcsin\left(\frac{1}{3}\right) + 2\pi k$ or $\theta = \pi - \arcsin\left(\frac{1}{3}\right) + 2\pi k$ for integers $k$. We find the two solutions here which lie in $[0,2\pi)$ to be $\theta = \arcsin\left(\frac{1}{3}\right)$ and $\theta = \pi - \arcsin\left(\frac{1}{3}\right)$.