Howdy math friends! In this post, we are going to learn about solving trigonometric equations algebraically. This will combine our knowledge of algebra and trigonometry into one beautiful question! For more on trigonometric functions and right triangle trigonometry check out this post here.
Solving trigonometric equations. Sound complicated? Well, you are correct, that does sound complicated. Is it complicated? Hopefully, you won’t fund it that way after you’ve seen this example. We are going to do this step by step in the following regents question:
How do I answer this question?
The questions want us to solve for x.
Step 1: Pretend this was any other equation and you wanted to solve for x. Thinking that way, we can move radical 2 to the other side.
Step 3: Now we need a value for x. This is where I turn to my handy dandy reference triangle. (This is a complete reference tool that you should
memorize know). For more on special triangles, check out this post here.
Step 5: Notice that cosine is positive in Quadrants I and IV. That means there are two values that x can be (one in each quadrant). We already have x=45º from Quadrant I. In order to get that other value in Quadrant IV, we must subtract 360º-45º=315º giving us our other value.
Does this make sense? Great! 🙂 Is it clear as mud? I have failed. But I have not given up (and neither should you). Ask more questions, look for the spots where you got lost, do more research and never give up! 🙂
Hopefully you enjoyed my short motivational speech. For more encouraging words and math, check out MathSux on the following websites! Sign up for FREE math videos, lessons, practice questions, and more. Thanks for stopping by and happy calculating! 🙂