1. Copyright © 2007 Pearson Education, Inc. Slide 9-1 6.2 Trigonometric Equations and Inequalities (I) Solving a Trigonometric Equation by Linear Methods ExampleSolve 2 sin x – 1 = 0 over the interval [0, 2  ). Analytic SolutionSince this equation involves the first power of sin x, it is linear in sin x.

2. Copyright © 2007 Pearson Education, Inc. Slide 9-2 6.2 Solving a Trigonometric Equation by Linear Methods Graphing Calculator Solution Graph y = 2 sin x – 1 over the interval [0, 2  ]. The x-intercepts have the same decimal approximations as

3. Copyright © 2007 Pearson Education, Inc. Slide 9-3 6.2 Solving Trigonometric Inequalities ExampleSolve for x over the interval [0, 2  ). (a)2 sin x –1 > 0 and (b) 2 sin x –1 < 0. Solution (a)Identify the values for which the graph of y = 2 sin x –1 is above the x-axis. From the previous graph, the solution set is (b)Identify the values for which the graph of y = 2 sin x –1 is below the x-axis. From the previous graph, the solution set is

4. Copyright © 2007 Pearson Education, Inc. Slide 9-4 6.2 Equations Solvable by Factoring ExampleSolve tan 2 x + tan x –2 = 0 over the interval [0, 2  ). SolutionThis equation is quadratic in term tan x. The solutions for tan x = 1 in [0, 2  ) are x = Use a calculator to find the solution to tan -1 (–2)  –1.107148718. To get the values in the interval [0, 2  ), we add  and 2  to tan -1 (–2) to get x = tan -1 (–2) +   2.03443936 and x = tan -1 (–2) + 2   5.176036589.

5. Copyright © 2007 Pearson Education, Inc. Slide 9-5 6.2 Solving a Trigonometric Equation by Factoring ExampleSolve sin x tan x = sin x over the interval [0°, 360°). Solution Caution Avoid dividing both sides by sin x. The two solutions that make sin x = 0 would not appear.

6. Copyright © 2007 Pearson Education, Inc. Slide 9-6 6.2 Solving a Trigonometric Equation Using the Quadratic Formula ExampleSolve cot x(cot x + 3) = 1 over the interval [0, 2  ). SolutionRewrite the expression in standard quadratic form to get cot 2 x + 3 cot x – 1 = 0, with a = 1, b = 3, c = –1, and cot x as the variable. Since we cannot take the inverse cotangent with the calculator, we use the fact that

7. Copyright © 2007 Pearson Education, Inc. Slide 9-7 6.2 Solving a Trigonometric Equation Using the Quadratic Formula The first of these, –.29400113018, is not in the desired interval. Since the period of cotangent is , we add  and then 2  to –.29400113018 to get 2.847591352 and 5.989184005. The second value, 1.276795025, is in the interval, so we add  to it to get another solution. The solution set is {1.28, 2.85, 4.42, 5.99}.

8. Copyright © 2007 Pearson Education, Inc. Slide 9-8 6.2Solving a Trigonometric Equation by Squaring and Trigonometric Substitution ExampleSolve over the interval [0, 2  ). SolutionSquare both sides and use the identity 1 + tan 2 x = sec 2 x.