1. Trigonometric Equations 5.5

2. To solve an equation containing a single trigonometric function: Isolate the function on one side of the equation. Solve for the variable. Equations Involving a Single Trigonometric Function

3. y = cosx x y 1 –1 y = 0.5 –4  2  –2  4  cos x = 0.5 has infinitely many solutions for –  < x <  y = cosx x y 1 –1 0.5 2  cos x = 0.5 has two solutions for 0 < x < 2  Trigonometric Equations

4. Solve the equation: 3 sin x  2  5 sin x  1. Solution The equation contains a single trigonometric function, sin x. Step 1Isolate the function on one side of the equation. We can solve for sin x by collecting all terms with sin x on the left side, and all the constant terms on the right side. 3 sin x  2  5 sin x  1 This is the given equation. 3 sin x  5 sin x  2  5 sin x  5 sin x – 1 Subtract 5 sin x from both sides. sin x  - 1/2 Divide both sides by  2 and solve for sin x.  2 sin x  1 Add 2 to both sides.  2 sin x  2   1 Simplify. Text Example

5. Solve the equation: 2 cos 2 x  cos x  1  0, 0  x  2 . The solutions in the interval [0, 2  ) are  / 3, , and 5  / 3. Solution The given equation is in quadratic form 2t 2  t  1  0 with t  cos x. Let us attempt to solve the equation using factoring. 2 cos 2 x  cos x  1  0 This is the given equation. (2 cos x  1)(cos x  1)  0 Factor. Notice that 2t 2 + t – 1 factors as (2t – 1)(2t + 1). cos x  1/2 2 cos x  1 cos x  1 Solve for cos x. 2 cos x  1  0 or cos x  1  0 Set each factor equal to 0. Text Example x   x  2  x  

6. Example Solve the following equation: Solution:

7. Example Solve the equation on the interval [0,2  ) Solution:

8. 8 sin  = 3(1  sin 2  ) Use the Pythagorean Identity. Rewrite the equation in terms of only one trigonometric function. Example: Solve 8 sin  = 3 cos 2  with  in the interval [0, 2π]. 3 sin 2  + 8 sin   3 = 0. A “quadratic” equation with sin x as the variable Therefore, 3 sin   1 = 0 or sin  + 3 = 0 (3 sin   1)(sin  + 3) = 0 Factor. Solutions: sin  = or sin  = -3 1 3  = sin  1 ( ) = 0.3398 and  = π  sin  1 ( ) = 2.8107. 1 3 1 3 Example: Solutions in an interval

9. Example Solve the equation on the interval [0,2  ) Solution:

10. Example Solve the equation on the interval [0,2  ) Solution: NO!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!