1. 1 Start Up Day 37 1.Simplify: 2, Verify:

2. SOLVING TRIGONOMETRIC EQUATIONS-DAY 37 OBJECTIVE : SWBAT SOLVE TRIGONOMETRIC EQUATIONS. EQ: How can we use trigonometric identities to help solve trig equations? What is the difference between solving for all solutions (General Solution) verses solving within an interval like from 0 to 2π?. HOMELEARNING : p. 411 #53, 57, 60, 63 & 73 ; p. 411 #51, 52, 55, 56, 58, 59, 64, 66, 69, 71 & 87

3. 3 FOCUS on ALGEBRA 1 st All trigonometric equations are NOT identities. Some equations will only be true at specific angles. To solve a trigonometric equation, use standard algebraic techniques such as collecting like terms and factoring. Your preliminary goal in solving a trigonometric equation is to isolate the trigonometric function in the equation. Sometimes it is helpful to replace the trigonometric function with just an ordinary variable of your choosing so that you can focus first on the algebra. For example, to solve the equation 2 sin x = 1, we might let X=sin x & re-write it as 2X=1 then we can see clearly that we should divide each side by 2 to obtain X = ½ OR, by replacing our original function:

4. 4 Back to TRIG: Where is sin x = ½? To solve for x, note that the equation has solutions x =  /6 and x = 5  /6 in the interval [0, 2  ).

5. 5 Infinitely many solutions? Moreover, because sin x has a period of 2 , there are infinitely many other solutions, which can be written as and where n is an integer.. This is known as the “General solution” “+ 2nπ” means “+ any number of full circles”

6. 6 Back to the “Circle of Life” Another way to show that the equation has infinitely many solutions is to identify the angles on the unit circle where this sine value occurs. Any angles that are co-terminal with  /6 or 5  /6 will also be solutions of the equation. When solving trigonometric equations, you should write your answer(s) using exact values rather than decimal approximations. Figure 5.7

7. 7 Example 2 – Collecting Like Terms Solve Solution: Begin by rewriting the equation so that sin x is isolated on one side of the equation. Let X = sin x (temporarily!) Define your temporary variable.. Add “x” to each side. Subtract from each side. RE- Write original equation. Divide each side by 2.

8. 8 The basics of Solving Trig Equations http://www.voki.com/pickup.php?scid=8888 701&height=267&width=200

9. 9 Trigonometry & the UNIT Circle! Now, Identify the angles where this sine value occurs. Because sin x has a period of 2 , first find all solutions in the interval [0, 2  ). (Once around the UNIT Circle) These solutions are x = 5  /4 and x = 7  /4. Finally, add multiples of 2  to each of these solutions to get the general form and where n is an integer. cont’d Re-write with the original trig function.. General solution

10. 10 Equations of Quadratic Type

11. 11 Equations of Quadratic Type Many trigonometric equations are of quadratic type ax 2 + bx + c = 0. Here are a couple of examples. Quadratic in sin x Quadratic in sec x 2 sin 2 x – sin x – 1 = 0 sec 2 x – 3 sec x – 2 = 0 2(sin x) 2 – sin x – 1 = 0 (sec x) 2 – 3(sec x) – 2 = 0 To solve equations of this type, factor the quadratic or, if this is not possible, use the Quadratic Formula.

12. 12 Example 3 – Factoring an Equation of Quadratic Type Find all solutions of 2 sin 2 x – sin x – 1 = 0 in the interval [0, 2  ). Solution: Begin by treating the equation as a quadratic & let X= sin x temporarily so that you can see that factoring is the key! 2 X 2 – X– 1 = 0 (2 X + 1)(X– 1) = 0 Re-Write original equation. Factor.

13. 13 Example 3 – Solution Setting each factor equal to zero, you obtain the following solutions in the interval [0, 2  ). 2 X+ 1 = 0 and X– 1 = 0 X= - ½ and X = 1 Once the Algebra is done, bring back the original trigonometric function and identify the angles where these values occur from 0 to 2π We don’t need to add “2nπ” to our solutions because the solution interval was limited to a specific interval. [0, 2  ). cont’d

14. 14 YOUR TURN: 1.Solve over the interval 2. Solve: 3. Solve:

15. 15 Using Inverse Functions and Identities

16. 16 Using trigonometric Identities and Inverse Functions In the next example, you will see how identities and inverse trigonometric functions can be used to solve an equation.

17. 17 Example 4 – Using Identities & Inverse Functions Solve sec 2 x – 2 tan x = 4. Solution: sec 2 x – 2 tan x = 4 1 + tan 2 x – 2 tan x – 4 = 0 tan 2 x – 2 tan x – 3 = 0 X 2 – 2 X– 3 = 0 (X– 3)(X+ 1) = 0 Notice that 2 different trig Functions are terms withing the original equation. Use an identity to rewrite one trig function in Terms of the other– the Pythagorean identity Is most appropriate for this one! Combine like terms and re-write with a temporary variable. Factor.

18. 18 Example 4 – Working on the solution. Setting each factor equal to zero, you obtain two solutions, X= 3 and X = -1 OR tan x = 3 tan x = –1 Now identify the angles where these values occur, if possible. Since a tangent value of 3 is does NOT occur at one of our special angles from the UNIT Circle, the best way to represent this solution is by using the inverse or arc function: x = arctan 3 cont’d

19. 19 Example 4 – A Final Solution Finally, because tan x has a period of , you obtain the general solution by adding multiples of  x = arctan 3 + n  and where n is an integer. You can use a calculator to approximate the value of arctan 3, if necessary. (radian mode) cont’d General solution

20. 20 Functions Involving Multiple Angles

21. 21 Functions Involving Multiple Angles The next example involves trigonometric functions of multiple angles of the forms sin ku and cos ku. To solve equations of these forms, first ignore the “ku”, re- write with a variable of your choosing, solve the equation algebraically for your variable, replace with the original trig function and solve the the angle like you would normally do. Lastly, consider these “angles” are really solutions for ku, so we must divide your result by k.

22. 22 Example 5 – Functions of Multiple Angles Solve 2 cos 3t – 1 = 0. Solution: Let X = cos 3t 2 X– 1 = 0 2 X= 1 X = ½ Re-Write original equation. Add 1 to each side. Divide each side by 2. Replace your original trig function..

23. 23 Example 5 – Solution Now identify all the angles within the interval [0, 2  ), where the cosine value is ½ You should know that this occurs at  /3 and 5  /3, so, in general, you have but this is only solved for “3t” –so we will need a little more algebra Dividing these results by 3, you obtain the general solution and where n is an integer. cont’d General solution