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6.2 Trigonometric Equations and Inequalities (I)

6.2 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.

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6.2 Trigonometric Equations and Inequalities (I)

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  1. 6.2 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.

  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. 6.2 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

  4. 6.2 Equations Solvable by Factoring Example Solve tan2x + 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.

  5. 6.2 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.

  6. 6.2 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 cot2x + 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. 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. 6.2 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 + tan2x = sec2x.

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