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# Class Note for MATH 111 with Professor Comeau at UA

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This 3 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Arizona taught by a professor in Fall. Since its upload, it has received 21 views.

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Date Created: 02/06/15

Math 111 Trigonometric Equation Problems Darin Comeau Section 4 Here are some additional problems where you solve trigonometric equations For each of the below equations nd all x that make the statement true Refer to the online notes for some worked examples 1 sinztanx sinz z E 027T Solution Subtracting sinx from both sides we have sinx tanm 7 sinx 0 sin xtan 7 1 0 sinz0 or tanz710 ln 0 27f sinx 0 for z 0 7139 27139 and tanm 1 for z 7r4 57r4 remember tanm is positive for z terminating in quadrants l and Ill Then our solutions are 2 tan2tanz720 Solution We recognize this as a quadratic in the variable tan x Then using the quadratic formula we get the roots of the above equation are 711 17472 7 nix67 7113 2 7 2 7 2 Therefore the roots of the quadratic equation are tanx 72tan z 1 These are the equations we now need to solve but we7ll rst write the quadratic in its factored form to see why we are solving them tanx tan2xtanz7 2 tanx71tan2 0 Since tanx is giving us our solutions we will nd all solutions in 7T27T2 there will be one from each equation and add integer multiples of 7139 the period of tangent tanx 1 is true for z 7T4 and tanx 72 taking arctan of both sides gives us z arctan72 Note in this example we only have to look for one solution to each equation since tanx is one to one on one cycle here 77r2 7r2 Then we add 7m to each solution to nd all solutions 7m arctan72 7m 3 cos2x cos x Solution We want to use a double angle identity for cos 2x that will leave us with factors of cos x so we use cos 2x 2 cos2 x 7 1 Substituting this in7 we have 2cos2x 71 cos 2cos2x icoss 71 0 We now factor using the quadratic formula we have 1i 174271 71i3 22 7 4 Therefore the solutions to the quadratic equation note these are not solutions to our problem are cos 17 712 COS l 2cos2z7cosx71cosz71 ltcosx 0 So we rst nd the solutions to cos x 1 in 07 27f7 which is only z 0 since this is the maximum of cos x The solutions to cos 712 in 07 27139 are x 27r3747r3 Then adding 27m to all our solutions7 we have 027m s 2quot27m 4 27m 4 4sincosx Solution We recognize the identity sin2 2sinxcos x and use this to rewrite the expression in terms of sin2x We rst divide each side by 2 2sinxcos wms sin2 Now we know how to nd the solutions to sin 6 2 the solutions in 07 27139 are 6 7r37 27r3 So letting 6 2x we get z 7T677T3 Now here is where we would normally add 27m to each solution however here the function that is giving us our solutions is sin27 not sin 7 and the period of sin2z is 7139 So we actually need to add integer multiples of 7139 instead g7m 7r 7m 5 cos 2x 2sinz 71 Solution This one is actually more dif cult than I had imagined As we7ll see the numbers we get in the end aren7t 77nice We7ll start by rewriting cos2x by a double angle identity We have three versions to choose one7 and the one we want is the one that involves only factors of sinx since those factors are already present in our equation cos2z 2sinx71 172sin2x 2sinx72 72sin2x 7 2sinz3 0 Now we use the quadratic formula to factor this 2ix227472372im7 Qi x i 1i 7 4 s1n 272 4 7 2 So we have 1 7 17 7 72 sin2 z 7 2 sinx 3 sinz Bf sinz 2f 0 For each factor7 we try to nd one solution using inverse sine However 1 W2 gt 1 is out of the range of sin x and thus there cannot be a solution to sinx 71 27 so the only solutions will come from the sinz factor well use inverse sine to nd one7 and for the other solution in 07 27f7 we actually need to use the identity sin7r 7 s sinx this is a consequence of the identity for Sine of a Difference Then all solutions will be arcsin 27m 7139 7 arcsin 27m If the last few steps of this example didnt make sense7 that7s ok I didnt realize the problem would be this complicated when l assigned it7 and you wont have anything this dif cult on the nal I would just make sure you understand how to factor equations that are quadratic in a trig function7 like in 273

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