1. Solving Trigonometric Equations Section 10-4

2. Acceleration due to gravity: g = 9.8 m/s 2. In physics, you learn that this is a constant, but in actuality, this value changes slightly based on your latitude. You can approximate g using the following formula, where  represents the degrees of latitude: g  9.78049(1 + 0.005288 sin 2  - 0.000006 sin 2 2  ) What is g in Chicago, which has a latitude of 42  N? http://www.forestwaterfalls.com/images/waterfall-23.JPG  9.81504 Does g get larger or smaller as you increase your latitude? Why?

3. Methods for Solving the Trigonometric Equation f(x) = g(x) Graphing a)Use a graphing calculator and graph y 1 = f(x) and y 2 = g(x) b)Use 2nd Calc Intersect to find the points of intersection Algebraically (usually the method of choice) a)Draw a quick sketch of the graphs y = f(x) and y = g(x) so you know how many solutions to look for. b)If the equation involves functions of 2x AND x, transform the functions of 2x into functions of x by using identities. c)If the equation involves functions of 2x ONLY, then it’s usually better to solve for 2x directly and then find x. d)Be careful not to lose (or gain) any roots!

4. Let’s review a little bit of what we used to do: 1)Solve 2cos 2 x - sinx = 1 for 2)Solve 2cos 2 x = cosx for 3)Solve √3cos2x = 3sin2x for

5. Solve cos2x = 1 - sinx for 0 ≤ x < 2π There are four points of intersection in the range we are given. Let’s graph the functions to see what we’re working with. y = 1 - sinx y = cos2x

6. Solve cos2x = 1 - sinx for 0 ≤ x < 2π Now that we know how many points to look for, let’s solve this problem algebraically. The equation contains a function of 2x and a function of just x, so we should convert the 2x function into something with functions of just x. So obviously we have to change cos2x into something else… but which version of cos2x should we use? cos2x = 1 - 2sin 2 x y = 1 - sinx y = cos2x

7. Solve cos2x < 1 - sinx for 0 ≤ x < 2π Now that we know where the two graphs intersect, we can study the positions of the graphs to determine when cos2x is less than 1 - sinx. What does that physically look like on a graph, for one function to be “less” than another? So we’re looking for when cos2x is below the graph of 1 - sinx. That happens when: and. y = 1 - sinx y = cos2x

8. Solve 3cos2x + cosx = 2 for 0 ≤ x < 2π What method would we have to use here to solve this equation algebraically? Turn cos2x into 2cos 2 x - 1 So now we have 3(2cos 2 x - 1) + cosx = 2, and you can solve that! 2π y = 2 - cosx y = 3cos2x

9. Solve 2sin2x = 1 for 0 ≤ x < 360º What method would we have to use here to solve this equation algebraically? Since we only have functions of 2x, and no functions of x, we can solve directly for 2x and then find x using what we know about the graph and the period of the function y = sin2x. This isn’t anything new. We solved equations like this in Chapter 8. In fact, #3 on your Chapter 8 test was to solve 2sin2x - 1 = 0. So… what’s x? 2π

10. Find the smallest positive root of sin2.8x = cosx Can you think of a way to solve this problem algebraically? We’re actually MUCH better off solving this problem on a graphing calculator. y = cosx y = sin2.8x 2π

11. Try these problems now: 1) cos2x - sinx > 0 2) cos2x = cosx 3) cos2x < 6cosx - 9