745 7.6 Trigonometric Equations Equations with Multiple Angles EXAMPLE 7 Solving an Equation Using a Double-Angle Identity Solve cos 2x = cos x over the interval 30, 2p2. SOLUTION First convert cos 2x to a function of x alone. Use the identity cos 2x = 2 cos2 x - 1 so that the equation involves only cos x. Then factor. cos 2x = cos x Original equation 2 cos2 x - 1 = cos x Cosine double-angle identity 2 cos2 x - cos x - 1 = 0 Subtract cos x. 12 cos x + 12 1cos x - 12 = 0 Factor. 2 cos x + 1 = 0 or cos x - 1 = 0 Zero-factor property cos x = - 1 2 or cos x = 1 Solve each equation for cos x. If we use the unit circle to analyze these results, we recognize that a radian- measured angle having cosine - 1 2 must be in quadrant II or III with reference angle p 3 . Another possibility is that it has a value of 1 at 0 radians. We can use Figure 33 to determine that solutions over the required interval are as follows. x = 2p 3 or x = 4p 3 or x = 0 The solution set is E0, 2p 3 , 4p 3 F . S Now Try Exercise 79. y x –1 –1 1 4P 3 P 3 P 3 2P 3 (– , ) 1 2 2 √3 (– , – ) 1 2 2 √3 0 (1, 0) 1 2 – Figure 33 EXAMPLE 8 Solving an Equation Using a Double-Angle Identity Solve 4 sin u cos u = 23 (a) over the interval 30°, 360°2 (b) for all solutions. SOLUTION (a) 4 sin u cos u = 23 Original equation 212 sin u cos u2 = 23 4 = 2 # 2 2 sin 2u = 23 Sine double-angle identity sin 2u = 23 2 Divide by 2. From the given interval 0° … u 6360°, the corresponding interval for 2u is 0° … 2u 6720°. Because the sine is positive in quadrants I and II, solutions over this interval are as follows. 2u = 60°, 120°, 420°, 480°, Reference angle is 60°. or u = 30°, 60°, 210°, 240° Divide by 2. The final two solutions for 2u were found by adding 360° to 60° and 120°, respectively, which gives the solution set 530°, 60°, 210°, 240°6.
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