SECTION 7.3 Trigonometric Equations 495 When the argument of the trigonometric function in an equation is a multiple of θ, the general formula is required to solve the equation. Now Work PROBLEM 29 Solving a Trigonometric Equation Involving a Double Angle Solve the equation: θ θ π ( ) = ≤ < sin 2 1 2 , 0 2 Solution EXAMPLE 4 In the interval π [ ) 0, 2 , the sine function equals 1 2 at π 6 and π5 6 . See Figure 27(a). In the interval π [ ) 0, 2 , the graph of θ ( ) = y sin 2 completes two cycles, and the graph of θ ( ) = y sin 2 intersects the graph of = y 1 2 four times. See Figure 27(b). So, there are four solutions of the equation θ ( ) = sin 2 1 2 in the interval π [ ) 0, 2 . To find the solutions, use θ2 in the general formula that gives all the solutions. θ π π θ π π = + = + k k k 2 6 2 or 2 5 6 2 an integer The argument is θ2 . θ π π θ π π = + = + k k 12 or 5 12 Divide by 2. Then θ π π π ( ) = + − = − 12 1 11 12 k 1 =− θ π π π ( ) = + − = − 5 12 1 7 12 θ π π π = + ⋅ = 12 0 12 k 0 = θ π π π = + ⋅ = 5 12 0 5 12 θ π π π = + ⋅ = 12 1 13 12 k 1 = θ π π π = + ⋅ = 5 12 1 17 12 θ π π π = + ⋅ = 12 2 25 12 k 2 = θ π π π = + ⋅ = 5 12 2 29 12 In the interval π [ ) 0, 2 , the solutions of θ ( ) = sin 2 1 2 are θ π θ π = = 12 , 5 12 , θ π = 13 12 , and θ π = 17 12 . The solution set is π π π π { } 12 , 5 12 , 13 12 , 17 12 . The graph of θ ( ) = y sin 2 intersects the graph of = y 1 2 at the points π ( ) 12 , 1 2 , π π ( ) ( ) 5 12 , 1 2 , 13 12 , 1 2 , and π ( ) 17 12 , 1 2 in the interval π [ ) 0, 2 . Check: Verify these solutions by graphing ( ) = Y x sin 2 1 and = Y 1 2 2 for π ≤ ≤x 0 2 . Figure 27 x y (21, 0) (0, 21) (0, 1) (1, 0) (x, ) (2x, ) p–– 6 1 – 2 1 – 2 (a) x2 1 y2 5 1 2u 5 5p ––– 6 2u 5 1 y y 5 21 2p p u (b) 1 –– 2 CAUTION In solving a trigonometric equation for θ θ π ≤ < , 0 2 , in which the argument is not θ (as in Example 4), you must write down all the solutions first and then list those that are in the interval π [ ) 0, 2 . Otherwise, solutions may be lost. j Solving a Trigonometric Equation Solve the equation: θ π θ π ( ) − = ≤ < tan 2 1, 0 2 Solution EXAMPLE 5 The period of the tangent function is π. In the interval π [ ) 0, , the tangent function has the value 1 when the argument is π 4 . Because the argument is θ π − 2 in the given equation, use it in the general formula that gives all the solutions. (continued)
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