Chapter 9: Trigonometric Identities and Equations (I) 9.2 Sum and Difference Identities 9.3 Further Identities 9.4 The Inverse Circular Functions 9.5 Trigonometric Equations and Inequalities (I) 9.6 Trigonometric Equations and Inequalities (II)

9.5 Trigonometric Equations and Inequalities (I) Solving a Trigonometric Equation by Linear Methods Example Solve 2 sin x – 1 = 0 over the interval [0, 2). Analytic Solution Since this equation involves the first power of sin x, it is linear in sin x.

9.5 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

9.5 Solving Trigonometric Inequalities Example Solve for x over the interval [0, 2). 2 sin x –1 > 0 and (b) 2 sin x –1 < 0. Solution 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

9.5 Solving a Trigonometric Equation by Factoring Example Solve 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.

9.5 Equations Solvable by Factoring Example Solve tan2 x + tan x –2 = 0 over the interval [0, 2). Solution This 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.

9.5 Solving a Trigonometric Equation Using the Quadratic Formula Example Solve cot x(cot x + 3) = 1 over the interval [0, 2). Solution Rewrite the expression in standard quadratic form to get cot2 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

9.5 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}.

9. 5. Solving a Trigonometric Equation by 9.5 Solving a Trigonometric Equation by Squaring and Trigonometric Substitution Example Solve over the interval [0, 2). Solution Square both sides and use the identity 1 + tan2 x = sec2 x.

9.4 The Inverse Sine Function Solving a Trigonometric Equation Analytically Decide whether the equation is linear or quadratic, so you can determine the solution method. If only one trigonometric function is present, solve the equation for that function. If more than one trigonometric function is present, rearrange the equation so that one side equals 0. Then try to factor and set each factor equal to 0 to solve. If the equation is quadratic in form, but not factorable, use the quadratic formula. Check that solutions are in the desired interval. Try using identities to change the form of the equation. It may be helpful to square each side of the equation first. If this is done, check for extraneous values.

9.4 The Inverse Sine Function Solving a Trigonometric Equation Graphically For an equation of the form f(x) = g(x), use the intersection-of-graphs method. For an equation of the form f(x) = 0, use the x-intercept method.