741 7.6 Trigonometric Equations y x 45° 225° 0° 180° (1, 0) (–1, 0) –1 1 (– , – ) 2 √2 2 √2 ( , ) 2 √2 2 √2 Figure 31 y x –1.1071 –1.1071 + 2P = 5.1760 –1.1071 + P = 2.0344 1 –1 –1 1 The solutions shown in blue represent angle measures, in radians, and their intercepted arc lengths on the unit circle. 5P 4 P 4 Figure 32 Zero-Factor Property Method EXAMPLE 2 Solving a Trigonometric Equation (Zero-Factor Property) Solve sin u tan u = sin u over the interval 30°, 360°2. SOLUTION sin u tan u = sin u Original equation sin u tan u - sin u = 0 Subtract sin u. sin u 1tan u - 12 = 0 Factor out sin u. sin u = 0 or tan u - 1 = 0 Zero-factor property tan u = 1 u = 0° or u = 180° u = 45° or u = 225° Apply the inverse function. See Figure 31. The solution set is 50°, 45°, 180°, 225°6. S Now Try Exercise 35. CAUTION Trying to solve the equation in Example 2 by dividing each side by sin u would lead to tan u = 1, which would give u = 45° or u = 225°. The missing two solutions are the ones that make the divisor, sin u, equal 0. For this reason, we avoid dividing by a variable expression. Quadratic Methods The equation au2 + bu + c = 0, where u is an algebraic expression, is solved by quadratic methods. The expression u may be a trigonometric function. EXAMPLE 3 Solving a Trigonometric Equation (Zero-Factor Property) Solve tan2 x + tan x - 2 = 0 over the interval 30, 2p2. SOLUTION tan2 x + tan x - 2 = 0 This equation is quadratic in form. 1tan x - 12 1tan x + 22 = 0 Factor. tan x - 1 = 0 or tan x + 2 = 0 Zero-factor property tan x = 1 or tan x = -2 Solve each equation. The solutions for tan x = 1 over the interval 30, 2p2 are x = p 4 and x = 5p 4 . To solve tan x = -2 over that interval, we use a calculator set in radian mode. We find that tan-11-22 ≈ -1.1071487. This is a quadrant IV number, based on the range of the inverse tangent function. However, because we want solutions over the interval 30, 2p2, we must first add p to -1.1071487 and then add 2p. See Figure 32. x ≈ -1.1071487 + p≈2.0344439 x ≈ -1.1071487 + 2p≈5.1760366 The solutions over the required interval form the following solution set. E p 4 , 5p 4 , 2.0344, 5.1760F (1111)1111* (111111111)111111111* Exact values Approximate values to four decimal places S Now Try Exercise 25.
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