1. Trigonometric Equations Solving for the angle (The second of two note days and a work day) (6.2)(2)

2. POD True/ false: The x-intercepts of y = sin 2x are ±πn/2.

3. POD True/ false: The x-intercepts of y = sin 2x are π/2 ±πn/2. There are couple of ways to do this. 1. Set sin 2x = 0, and solve. 2. Consider the x-intercepts of the graph of y = sin x. We’ve changed something in the equation—how does that change the graph? What x-scale could help see it?

4. Pick up from last time Solve by factoring. What are the four steps again?

5. Pick up from last time Step one: Isolate trig functions by factoring.

6. Pick up from last time Start step 2: use inverse trig functions.

7. Pick up from last time Step two: Find angles in one rotation. Note: the inverse trig function for the last one gives us a value outside the interval 0 ≤ θ ≤ 2π. If we use that tool, we have to build on it.

8. Pick up from last time Step three: Find the general solution (all angles). We can combine those last four into this. Or this.

9. Solving for the variable Solve for u. (Note, this is not 2 csc 4 u.) You could replace the 2u with x or θ for now, if it helps.

10. Solving for the variable Step one: Isolate trig functions by factoring. The second factor does not provide a meaningful solution.

11. Solving for the variable Step two: Find angles in one rotation. Again, we build off of the angle we get for the negative sine value.

12. Solving for the variable Step three: Find all angles. Combined, what would the statement be?

13. Solving for the variable Step three: Find all angles. Combined, what would the statement be? We can use this to solve for u. We can also use the separate statements to solve for u. We’ll get the same answer.

14. Solving for the variable Step four: Solve for u with all statements. This is a mouthful. What’s the easier way to present it?

15. Solving for the variable An elegant combination: We could get this using that single statement from before.

16. Approximating solutions Group and factor this to start. Work in degrees for a switch.

17. Approximating solutions Step one: Group and factor this to start.

18. Approximating solutions Step two: Find angles in one rotation. Once again, we have a factor that does not provide a meaningful solution.

19. Approximating solutions Step three: Find all angles.

20. Using a graph Graph to find the roots of this equation in one rotation. What patterns do you see? What x-scale could help to see the intercepts?

21. Using a graph Graph to find the roots of this equation in one rotation. Now graph it -2π to 2π. What do you see?

22. Using a graph What about this one? How does it compare to the one before?

23. Using a graph What about this one? x =.171, 1.24. What do you see here? If we rewrote the equation to set one side to 0, what would we have done to the graph?

24. Application From example 9, page 476. What do the various elements of this equation signify?

25. Application From example 9, page 466. How many days have more than 10.5 hours? What’s a fast way to find out?

26. Application From example 9, page 466. How many days have more than 10.5 hours? Graph it first on calculators and calculate intersections.

27. Application From example 9, page 466. How many days have more than 10.5 hours? Graph it first on calculators and calculate intersections. x = 48.6 and 291.9 Now, algebraically.

28. Application How many days have more than 10.5 hours? The expression is actually the angle.

29. Application The expression is actually the angle. Substitute θ for this expression, and solve for θ. sinθ = -.5 θ = -π/6 ≈ -.52θ = π/6 + π ≈ 3.67 Un-substitute and solve for t.

30. Application Un-substitute and solve for t. Find the difference between the days to answer the question (finally). Graphing was easier…