SECTION 7.3 Trigonometric Equations 497 Solving a Trigonometric Equation Quadratic in Form Solve the equation: θ θ θ π − + = ≤ < 2sin 3sin 1 0, 0 2 2 Solution EXAMPLE 7 This equation is a quadratic equation in θ sin that can be factored. θ θ − + = 2sin 3sin 1 0 2 θ − + = = x x x 2 3 1 0, sin 2 θ θ ( )( ) − − = 2sin 1 sin 1 0 x x 2 1 1 0 ( )( ) − − = θ θ − = − = 2sin 1 0orsin 1 0 Use the Zero-Product Property. θ θ = = sin 1 2 or sin 1 Solving each equation in the interval π [ ) 0, 2 yields θ π θ π θ π = = = 6 5 6 2 The solution set is π π π { } 6 , 2 , 5 6 . Figure 29 shows the solutions using Desmos. Figure 29 Now Work PROBLEM 63 4 Solve Trigonometric Equations Using Fundamental Identities Often when a trigonometric equation contains more than one trigonometric function, identities can be used to obtain an equivalent equation that contains only one trigonometric function. Solving a Trigonometric Equation Using Identities Solve the equation: θ θ θ π + = ≤ < 3cos 3 2 sin , 0 2 2 Solution EXAMPLE 8 The equation contains a sine and a cosine. However, using the Pythagorean Identity, θ θ + = sin cos 1, 2 2 the equation is transformed into an equivalent one containing only cosines. θ θ + = 3cos 3 2 sin2 θ θ ( ) + = − 3cos 3 2 1 cos2 sin 1 cos 2 2 θ θ = − θ θ + = − 3cos 3 2 2 cos2 Quadratic in cosθ θ θ + + = 2 cos 3cos 1 0 2 θ θ ( )( ) + + = 2 cos 1 cos 1 0 Factor. θ θ + = + = 2 cos 10 or cos 1 0 Use the Zero-Product Property. θ θ = − = − cos 1 2 or cos 1 Solving each equation in the interval π [ ) 0, 2 yields θ π θ π θ π = = = 2 3 4 3 The solution set is π π π { } 2 3 , , 4 3 . Check: Graph = + Y x 3cos 3 1 and π = ≤ ≤ Y x x 2sin , 0 2 , 2 2 and find the points of intersection. How close are your approximate solutions to the exact solutions? Now Work PROBLEM 79 ( ) = − + y x x 2 sin 3sin 1 2
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