Chapter 7 Part 5: Solving Trigonometric Equations

Welcome back to Professor Baker's Math Class! Today, we delved deeper into the world of trigonometry, focusing on solving trigonometric equations. We explored various techniques for finding all possible solutions, both general solutions and solutions within a specific interval. Let's recap the key concepts covered in class.

Finding General Solutions

When solving trigonometric equations, remember that trigonometric functions are periodic. This means they repeat their values at regular intervals. Therefore, we often have infinitely many solutions. To express these solutions generally, we use the following approach:

  • Isolate the trigonometric function: Get the trigonometric function (e.g., $\cos(x)$, $\sin(x)$, $\tan(x)$) by itself on one side of the equation.
  • Find the principal solutions: Determine the solutions within one period of the function (e.g., $[0, 2\pi)$ for sine and cosine).
  • Express the general solution: Add integer multiples of the period to the principal solutions. For example, if $\cos(y) = 0$, then $y = \frac{\pi}{2} + k\pi$, where $k$ is an integer.

Example: Find all solutions to $\cos(2x) = 0$.

  1. We have $\cos(2x) = 0$.
  2. Therefore, $2x = \frac{\pi}{2} + k\pi$.
  3. Dividing by 2, we get $x = \frac{\pi}{4} + \frac{k\pi}{2}$.

Finding Solutions in an Interval

Sometimes, we're only interested in solutions within a specific interval, such as $[0, 2\pi)$. In this case, we find the general solution first, then determine which values of $k$ give us solutions within the desired interval.

Example: Find all solutions to $2\cos(\frac{2x}{3}) - 1 = 0$ in the interval $[0, 2\pi)$.

  1. Isolate the cosine function: $\cos(\frac{2x}{3}) = \frac{1}{2}$.
  2. Find the angles where cosine equals $\frac{1}{2}$: $\frac{2x}{3} = \frac{\pi}{3}$ and $\frac{2x}{3} = \frac{5\pi}{3}$.
  3. Solve for x: $x = \frac{\pi}{2}$ and $x = \frac{5\pi}{2}$. However, $\frac{5\pi}{2}$ is outside the interval $[0, 2\pi)$.
  4. Considering periodicity, $\frac{2x}{3} = \frac{\pi}{3} + 2k\pi$ and $\frac{2x}{3} = \frac{5\pi}{3} + 2k\pi$. Solving for x gives $x = \frac{\pi}{2} + 3k\pi$ and $x = \frac{5\pi}{2} + 3k\pi$. The only solution in the interval $[0, 2\pi)$ is $x = \frac{\pi}{2}$.

Using Trigonometric Identities

Trigonometric identities are crucial for simplifying equations. Some common identities include:

  • $\sin^2(x) + \cos^2(x) = 1$
  • $\cos(2x) = \cos^2(x) - \sin^2(x) = 1 - 2\sin^2(x) = 2\cos^2(x) - 1$

Example: Solve $\cos(x) = -\sin^2(x) - 1$ in the interval $[0, 2\pi)$.

  1. Use the identity $\sin^2(x) = 1 - \cos^2(x)$: $\cos(x) = -(1 - \cos^2(x)) - 1$
  2. Simplify: $\cos(x) = \cos^2(x) - 2$
  3. Rearrange: $\cos^2(x) - \cos(x) - 2 = 0$
  4. Factor: $(\cos(x) - 2)(\cos(x) + 1) = 0$
  5. Solve: $\cos(x) = 2$ (no solution) or $\cos(x) = -1$, which gives $x = \pi$.

Real-World Application

Trigonometric equations can model real-world situations. Consider the height of a person on a Ferris wheel, $h(t) = 18.5 - 16.4\cos(\frac{2\pi}{5}t)$. Setting $h(t)$ to a particular height (e.g., 28 meters) allows us to solve for the time $t$ when the person is at that height.

Keep practicing, and you'll become a pro at solving trigonometric equations! Remember to utilize your unit circle and trigonometric identities. See you in the next class!