740 CHAPTER 7 Trigonometric Identities and Equations Earlier we studied trigonometric equations that were identities. We now consider trigonometric equations that are conditional. These equations are satisfied by some values but not others. 7.6 Trigonometric Equations ■ Linear Methods ■ Zero-Factor Property Method ■ Quadratic Methods ■ Trigonometric Identity Substitutions ■ Equations with Half-Angles ■ Equations with Multiple Angles ■ Applications Linear Methods The most basic trigonometric equations are solved by first using properties of equality to isolate a trigonometric expression on one side of the equation. EXAMPLE 1 Solving aTrigonometric Equation (Linear Methods) Solve the equation 2 sin u + 1 = 0 (a) over the interval 30°, 360°2 (b) for all solutions. ALGEBRAIC SOLUTION (a) Because sin u is to the first power, we use the same method as we would to solve the linear equation 2x + 1 = 0. 2 sin u + 1 = 0 Original equation 2 sin u = -1 Subtract 1. sin u = - 1 2 Divide by 2. To find values of u that satisfy sin u = - 1 2, we observe that u must be in either quadrant III or quadrant IV because the sine function is negative only in these two quadrants. Furthermore, the reference angle must be 30°. The graph of the unit circle in Figure 29 shows the two possible values of u. The solution set is 5210°, 330°6. GRAPHING CALCULATOR SOLUTION (a) Consider the original equation. 2 sin u + 1 = 0 We can find the solution set of this equation by graphing the function y1 = 2 sin x + 1 and then determining its zeros. Because we are finding solutions over the interval 30°, 360°2, we use degree mode and choose this interval of values for the input x on the graph. The screen in Figure 30(a) indicates that one solution is 210°, and the screen in Figure 30(b) indicates that the other solution is 330°. The solution set is 5210°, 330°6, which agrees with the algebraic solution. y x 330° 210° 30° 30° 1 –1 –1 1 1 2 – 1 2 (– , – ) 2 √3 1 2 ( , – ) 2 √3 Figure 29 (b) Because the graph of y1 = 2 sin x + 1 repeats the same y-values every 360°, all solutions are found by adding integer multiples of 360° to the solutions found in part (a). See the algebraic solution. S Now Try Exercises 15 and 47. −4 4 0° 360° −4 4 0° 360° Degree mode (b) Figure 30 Degree mode (a) (b) To find all solutions, we add integer multiples of the period of the sine function, 360°, to each solution found in part (a). The solution set is written as follows. 5210° + 360°n , 330° + 360°n , where n is any integer6
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