Algebra & Trigonometry

743 7.6 Trigonometric Equations EXAMPLE 5 Solving aTrigonometric Equation (Squaring) Solve tan x + 23 = sec x over the interval 30, 2p2. SOLUTION We must rewrite the equation in terms of a single trigonometric function. Because the tangent and secant functions are related by the identity 1 + tan2 x = sec2 x, square each side and express sec2 x in terms of tan2 x. Atan x + 23 B 2 = 1sec x22 Square each side. tan2 x + 223 tan x + 3 = sec2 x 1x + y22 = x2 + 2x y + y2 tan2 x + 223 tan x + 3 = 1 + tan2 x Pythagorean identity 223 tan x = -2 Subtract 3 + tan2 x. tan x = - 12 3 Divide by 223. tan x = - 23 3 Rationalize the denominator. Solutions of tan x = - 23 3 over 30, 2p2 are 5p 6 and 11p 6 . These possible, or proposed, solutions must be checked to determine whether they are also solutions of the original equation. CHECK Don’t forget the middle term. tan x + 23 = sec x Original equation tan a 5p 6 b + 23≟sec a 5p 6 b - 23 3 + 323 3 ≟- 223 3 223 3 = - 223 3 False tan a 11p 6 b + 23≟sec a 11p 6 b - 23 3 + 323 3 ≟2 23 3 223 3 = 223 3 ✓ True As the check shows, only 11p 6 is a solution, so the solution set is E 11p 6 F . S Now Try Exercise 45. −4 4 0 2p Radian mode The graph shows that on the interval 30, 2p2, the only zero of the function y = tan x + 23 - sec x is 5.7595865, which is an approximation for 11p 6 , the solution found in Example 5. Solving aTrigonometric 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.

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