8.5 Solving More Difficult Trigonometric Equations

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8.5 Solving More Difficult Trigonometric Equations Objective To use trigonometric identities or technology to solve more difficult trigonometric equations. x y [Solution] Take the fourth root of both sides to obtain: cos(2u )= 23From the unit circle, the solutions for 2u are 2u = + k, k any integer. 6 Example 1: To find all solutions of cos4(2u ) = . 9 16 1 6 - 6 x = - 23x = 23 Answer: u = + k , for k is any integer. 12t2tExample 2: Find all solutions of the trigonometric equation: tan2 u + tan u = 0. The solutions for tan u = 0 are the values u = k, for k is any integer. Therefore, tanu = 0 or tanu = -1. tan2u + tanu = 0 Original equation tanu (tanu +1) = 0 Factor. The solutions for tan u = -1 are u = - + k, for k is any integer. 4 Some trigonometric equations can be transformed into equations that have quadratic form. Example 3: The trigonometric equation 2 cos2u 3 sinu 3 = 0 . 2 sin2u + 3 sinu + 1 = 0 implies that u = -/2 + 2k, from sin u = -1 (2 sinu + 1)(sinu + 1) = 0. Therefore, 2 sinu + 1 = 0 or sinu + 1 = 0. Solutions: u = - + 2k and u = + 2k, from sin u = - 6 7 6 1 2 It follows that sinu = - or sinu = -1. 1 2 [Solution] Use the Pythagorean identity: 2 (1 sin2u ) 3 sinu 3 = 0 . Steps for Solving: Isolate the Trigonometric function. Then solve for the angle exactly if the ratio represents known special triangle ratios. Or solve for the angle approximately using the appropriate inverse trigonometric function. Complete the List of Solutions: If you are not restricted to a specific interval and are asked to give a complete list of solutions (general solution), then remember that adding on any integer multiple of 2 represents a co-terminal angle with the equivalent trigonometric ratio. 8 sinu = 3(1 sin2u ) Use the Pythagorean Identity. [Solution] Rewrite the equation in terms of only one trigonometric function. Example 4: Solve 8 sin u = 3 cos2 u with u in the interval [0, 2]. 3 sin2u + 8 sinu 3 = 0. A quadratic equation with sin x as the variable Therefore, 3 sinu 1 = 0 or sinu + 3 = 0 (3 sinu 1)(sinu + 3) = 0 Factor. u = arcsin( ) ~ 0.3398 and u = arcsin( ) ~ 2.8107. 1 3 1 3 Solutions: sinu = or sinu = -3 1 3 Example 5: 5cos2u + cos u 3 = 0 for 0 u . This is the range of the inverse cosine function. The solutions are: u = cos 1(0.6810249 ) = 0.8216349 and u = cos 1(0.8810249) = 2.6488206 Therefore, cosu = 0.6810249 or 0.8810249. Use the calculator to find values of u in 0 u . [Solution] The equation is quadratic. Let u = cosu and solve 5u2 + u 3 = 0. u = -1 = 0.6810249 or -0.8810249 61 10 Example 6: Solve a trigonometric equation by factoring. tanxsin x = sin x tan xsin x sin x = 0 sin x tanx 1( )=0tan 1 0 or sin 0 x x = =2452 24 4(2 1)4x kx k kkt tt tt t tt t= += + = + += + +[Solution] tan 1 or sin 0 x x = =x kt =4x kt t = +Example 7: Solve tan2x 3tanx 4 = 0 in 0 x 2t. [Solution] Let u = tan x, then x2 3x 4 = 0 (x 4)(x + 1) = 0 x 4 = 0 or x + 1 = 0 x = 4 or x = 1 tan x = 4 or tan x = 1 x = arctan4 or x = 3t/4 (x = 76 or x = 135) x = t + arctan4 or x = 7t/4 ( x = 256 or x = 315) Example 8: Solve. 3sinx + 4 = 1/sinx in 0 x 2t . [Solution] Let x = sin x, then 3x + 4 = 1/x x(3x + 4) = x(1/x) 3x2 + 4x = 1 (3x 1)(x + 1) = 0 3x 1= 0 or x + 1 = 0 x = 1/3 or x = 1 sin x = 1/3 or sin x = 1 x = arcsin(1/3) or x = 3t/2 ( x = 19 or x = 270) x = t arcsin(1/3) ( x = 161) 2 2 Solve sin sin cos for 0 2 . x x x x t = s < Example 9:2 2sin sin cos x x x =( )2 2sin sin 1 sin x x x = 22sin sin 1 0 x x =( )( )2sin 1 sin 1 0 x x + =2sin 1 0 sin 1 0 x x + = =1sin sin 12x x = =7 11,6 6 2x xt t t= =[Solution] Solve sin tan 3sin . x x x = Example 10:sin tan 3sin x x x =sin tan 3sin =0 x x x ( )sin tan 3 0 x x =sin 0 tan 3 0 x x = =0, x t =tan 3 x =arctan3, arctan3 x t = +[Solution] Particular solutions are 2 x k t =arctan3 x kt = +General solutions are 2 x k t t = +x k t =Example 11: Solve 2 cos x + sec x = 0 [Solution] ( )212cos 0cos12cos 1 0cosxxxx+ =+ =21cos ,2x =1cos2x = is not a real number, thus the equation has no solution. 10cos x =Therefore, 22cos 1 0 x + =10cos x =or However, So, 22cos 1 x = Example 12: Solve cos x + 1 = sin x in [0, 2t] [Solution] 3, or 2 2x xt t t = =2cos 1 = 1 cos u u + Since 1 cos 1, 0 cos 1 2 u u s s s + s2Thus sin 0 sin 1 cos u u u > = 2 2cos 2cos 1 1 cos x x u + + = 22cos 2cos 0 x x + =2cos (cos 1) 0 x x + =2cos 0 or cos 1 0 x x = + =cos 0 or cos 1 x x = = Since we squared the original equation we have to check our answer. The only solutions are t/2 and t. Like solving algebraic equation, once we squared the original trigonometric equation, it may generate some extraneous solution. We need to check the solutions. Book Example 5. Solve 2sin cos 1 for 0 < 360 . u u u = + s 2sin cos 1 u u = +2 2 2sin 1 cos sin 1 cos u u u u = = ( )22 1 cos cos 1 u u = +( ) ( )2222 1 cos cos 1 u u ( = + ( ( )2 24 1 cos cos 2cos 1 u u u = + +2 24 4cos cos 2cos 1 u u u = + +20 5cos 2cos 3 u u = + ( )( )5cos 3 cos 1 0 u u + =5cos 3 0 cos 1 0 u u = + =3cos cos 15u u = = 53.3 ,306.9 180 u u = = Since we squared the original equation we have to check our answer. 2sin53.1 cos53.1 1 = +1.6 1.6 =2sin180 cos180 1 = +0 0 =2sin306.9 cos306.9 1 = +1.6 1.6 =the only solutions are 53.1 and 180 Book Example 5. Solve 2sin cos 1 for 0 < 360 . u u u = + s 2sin cos 1 u u = +Since 1 cos 1, 0 cos 1 2 u u s s s + sThen 2sin cos 1 0 u u = + >( ) ( )2222 1 cos cos 1 u u ( = + ( ( )2 24 1 cos cos 2cos 1 u u u = + +2 24 4cos cos 2cos 1 u u u = + +20 5cos 2cos 3 u u = + ( )( )5cos 3 cos 1 0 u u + =5cos 3 0 cos 1 0 u u = + =3cos cos 15u u = = 53.3 ,306.9 180 u u = = Since we squared the original equation we have to check our answer. the only solutions are 53.1 and 180 2Thus sin 0 sin 1 cos u u u > = 2sin53.1 cos53.1 1 = +1.6 1.6 =2sin180 cos180 1 = +0 0 =2sin306.9 cos306.9 1 = +1.6 1.6 =a) Reference Angle: Therefore: b) The equation cannot be factored. Therefore, use the quadratic equation to find the roots: Reference Angles: sin u =13tanu = b b aca242tan( ) ( ) ( )( )( )u = 1 1 4 3 12 324 1 sin sin u u =3 1 sinu =u = 0 3398 .u = 0 3398 2 8018 . . and3 1 02tan tan u u =tan . tan . u u = = 0 43 0 76 oru = 0 4061 . u = 0 6499 .u = 2 7355 5 877 0 6499 3 7915 . , . , . , .Using a Graphing Calculator to Solve Trigonometric Equations 0 s u s 2t u y = 3sin -15.2.13 3tan2u tanu 1 = 0Using a Graphing Calculator to Solve Trigonometric Equations Therefore, u = 0.654, 2.731, 3.796, and 5.873 . Remark Solving trigonometric equations is a long lasting task through out the entire trigonometry. After we learned the sum and difference of angles, double angles, triple angles, we will be able solve some much more difficult trigonometric equations. Assignment P. 326 #1 22