763 CHAPTER 7 Test Prep Concepts Examples Solve tan u + 23 = 223 over the interval 30°, 360°2. tan u + 23 = 223 Original equation tan u = 23 Subtract 23. u = 60° Definition of inverse tangent Another solution over 30°, 360°2 is u = 60° + 180° = 240°. The solution set is 560°, 240°6. Solve 2 cos2 x = 1 for all solutions. 2 cos2 x = 1 Original equation 2 cos2 x - 1 = 0 Subtract 1. cos 2x = 0 Cosine double-angle identity 2x = p 2 + 2np and 2x = 3p 2 + 2np Add integer multiples of 2p. x = p 4 + np and x = 3p 4 + np Divide by 2. The solution set, where p is the period of cos 2x, is E p 4 + np, 3p 4 + np, where n is any integerF. 7.6 Trigonometric Equations Solving a Trigonometric Equation 1. Decide whether the equation is linear or quadratic in form in order to determine the solution method. 2. If only one trigonometric function is present, solve the equation for that function. 3. If more than one trigonometric function is present, rewrite the equation so that one side equals 0. Then try to factor and apply the zero-factor property. 4. If the equation is quadratic in form, but not factorable, use the quadratic formula. Check that solutions are in the desired interval. 5. Try using identities to change the form of the equation. It may be helpful to square each side of the equation first. In this case, check for extraneous solutions. 7.7 Equations Involving Inverse Trigonometric Functions We solve equations of the form y = ƒ1x2, where ƒ1x2 involves a trigonometric function, using inverse trigonometric functions. Techniques introduced in this section also show how to solve equations that involve inverse functions. Solve y = 2 sin 3x for x, where x is restricted to the interval C - p 6 , p 6 D . y = 2 sin 3x Original equation y 2 = sin 3x Divide by 2. 3x = arcsin y 2 Definition of arcsine x = 1 3 arcsin y 2 Multiply by 1 3 . Solve 4 tan-1 x = p. 4 tan-1 x = p Original equation tan-1 x = p 4 Divide by 4. x = tan p 4 Definition of arctangent x = 1 Evaluate. The solution set is 516.
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