Trigonometric equations: finding every solution in the range

Trigonometric equations: finding every solution in the range

Solving trigonometric equations goes beyond finding one value. Many students, even in DP2, will find the principal value and stop. This often costs them marks on IB exams. I want to show you how to systematically find all solutions within a given range, a skill crucial for both AA and AI courses, at both SL and HL.

My classroom experience tells me this is a common trap. We spend time mastering the unit circle and graphs of trigonometric functions, yet when it comes to solving equations, students forget that periodicity means multiple solutions exist. This article will break down the process step by step, ensuring you account for every possible solution.

Understanding the Periodic Nature

The core issue is periodicity. Functions like $\sin x$, $\cos x$, and $\tan x$ repeat their values. This means if $\sin x = k$ has one solution, it actually has infinitely many. Your job in the IB exam is to find the ones that fit within a specified interval, often $[0, 2\pi]$ or $[- \pi, \pi]$.

Let's consider $\sin x = 0.5$. Your calculator gives you $x = \frac{\pi}{6}$ (or $30^\circ$). This is the principal value. However, if you look at the unit circle, another angle in $[0, 2\pi]$ has the same sine value: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$. Similarly, $\cos x = 0.5$ gives $x = \frac{\pi}{3}$. The other solution in $[0, 2\pi]$ is $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$. For $\tan x = 1$, your calculator gives $x = \frac{\pi}{4}$. The other solution in $[0, 2\pi]$ is $x = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$.

Step-by-Step Approach for Simple Equations

I teach my students a four-step method for solving equations like $\sin(kx + c) = N$ or $\cos(kx + c) = N$. This is applicable to both SL and HL.

Step 1: Isolate the trigonometric function

Before you do anything else, make sure the trigonometric function is by itself. For example, if you have $2\sin x - 1 = 0$, rearrange it to $\sin x = \frac{1}{2}$. If it's $3\tan(2x) + \sqrt{3} = 0$, make it $\tan(2x) = -\frac{\sqrt{3}}{3}$.

Step 2: Find the principal value (and its pair)

Use your calculator (in radian mode, unless specified otherwise) to find the principal value. Let's call it $\alpha$. This is typically the value in $(-\frac{\pi}{2}, \frac{\pi}{2})$ for $\arcsin$, $[0, \pi]$ for $\arccos$, and $(-\frac{\pi}{2}, \frac{\pi}{2})$ for $\arctan$.

• For $\sin \theta = N$: Solutions are $\theta = \alpha$ and $\theta = \pi - \alpha$.
• For $\cos \theta = N$: Solutions are $\theta = \alpha$ and $\theta = 2\pi - \alpha$ (or $-\alpha$).
• For $\tan \theta = N$: Solutions are $\theta = \alpha$ and $\theta = \pi + \alpha$. (Remember $\tan$ has a period of $\pi$.)

These are the fundamental solutions within a $2\pi$ or $\pi$ cycle.

Step 3: Account for periodicity and the given range

This is where most students falter. You have the fundamental solutions. Now you need to add or subtract multiples of the period to find other solutions within the specified range.

For $\sin \theta = N$ and $\cos \theta = N$, the period is $2\pi$. So, if $\alpha$ is a solution, then $\alpha \pm 2\pi, \alpha \pm 4\pi, \dots$ are also solutions. For $\tan \theta = N$, the period is $\pi$. So, if $\alpha$ is a solution, then $\alpha \pm \pi, \alpha \pm 2\pi, \dots$ are also solutions.

If the equation is of the form $\sin(kx + c) = N$, then you need to adjust the range for the argument. If the original range for $x$ is, say, $[0, 2\pi]$, then for $kx+c$, the range becomes $[k(0)+c, k(2\pi)+c]$. This transformation of the range is critical. Let $u = kx+c$. Solve for $u$ in its transformed range, then solve for $x$.

Step 4: Check your solutions

Plug your solutions back into the original equation to ensure they work. Also, make sure every solution falls within the given range for $x$. I always encourage my students to write down all solutions then systematically discard those outside the range.

Example: $\cos(2x) = -\frac{1}{2}$ for $x \in [-\pi, \pi]$ (AA SL/HL, AI SL/HL)

1. The function is already isolated.

2. Find principal values for $\cos \theta = -\frac{1}{2}$. The principal value using $\arccos$ is $\theta = \frac{2\pi}{3}$. The other solution in $[0, 2\pi]$ is $\theta = 2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}$.

3. Account for periodicity and the transformed range. Let $u = 2x$. The range for $x$ is $[-\pi, \pi]$. So, the range for $u = 2x$ is $[2(-\pi), 2(\pi)] = [-2\pi, 2\pi]$. We need solutions for $\cos u = -\frac{1}{2}$ in $[-2\pi, 2\pi]$.

Our fundamental solutions are $u = \frac{2\pi}{3}$ and $u = \frac{4\pi}{3}$.

Now, add/subtract $2\pi$ (the period of cosine): $u_1 = \frac{2\pi}{3}$ (in range) $u_2 = \frac{4\pi}{3}$ (in range) $u_3 = \frac{2\pi}{3} - 2\pi = \frac{2\pi - 6\pi}{3} = -\frac{4\pi}{3}$ (in range) $u_4 = \frac{4\pi}{3} - 2\pi = \frac{4\pi - 6\pi}{3} = -\frac{2\pi}{3}$ (in range) If we added $2\pi$ to $u_1$ or $u_2$, we'd get values outside $2\pi$. If we subtracted $2\pi$ from $u_3$ or $u_4$, we'd get values outside $-2\pi$. So, the solutions for $u$ are $\{-\frac{4\pi}{3}, -\frac{2\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}\}$.

Now, substitute $u = 2x$ back and solve for $x$: $2x = -\frac{4\pi}{3} \Rightarrow x = -\frac{2\pi}{3}$ $2x = -\frac{2\pi}{3} \Rightarrow x = -\frac{\pi}{3}$ $2x = \frac{2\pi}{3} \Rightarrow x = \frac{\pi}{3}$ $2x = \frac{4\pi}{3} \Rightarrow x = \frac{2\pi}{3}$

4. Check solutions. All values $\{-\frac{2\pi}{3}, -\frac{\pi}{3}, \frac{\pi}{3}, \frac{2\pi}{3}\}$ are within the original range $x \in [-\pi, \pi]$.

This systematic approach, particularly the transformation of the range for the argument, is what distinguishes a complete solution from an incomplete one. I see many students rush this step, leading to missed solutions.

Equations with Identities (HL only for some, AA SL/HL)

Sometimes you need to use trigonometric identities to simplify the equation before solving. For example, $2\sin^2 x - \cos x = 1$. Here, you can replace $\sin^2 x$ with $1 - \cos^2 x$ to get a quadratic in $\cos x$. Or, $\sin(2x) + \sin x = 0$ requires using the double angle identity $\sin(2x) = 2\sin x \cos x$. These are common in Paper 1 AA SL and Paper 2 AI SL, but especially critical for HL.

Remember your Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$. Double angle identities: $\sin(2\theta) = 2\sin \theta \cos \theta$, $\cos(2\theta) = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta$. These are provided in your formula booklet. Know when and how to use them.

Final Thoughts

Mastering trigonometric equations is not just about memorizing formulas; it's about understanding the underlying periodic nature of the functions. Always work in radians unless degrees are explicitly specified. Be meticulous with your range transformations and don't assume your calculator gives you all solutions.

Practice is key. Work through examples, draw unit circles, and check your answers. If you're struggling, revisit the basics of the unit circle and graphs. This skill is foundational for calculus applications in both AA and AI courses. My study notes and flashcards can help reinforce these concepts. Keep practicing, and you'll find every solution every time.