26. What is the solution for the equation 29. It is the below? inc u = (sin r)(sin² 2u) Ce G be (a) cot 2u: = I - sin COS I + C (b) sec 2u cot 2u = sin r cos r -x+C (c) 2 cot 2u = sin r cos r- r +c (d) cot 2u sin 2r - x+ C %3| (e) None of the above 30

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.4: Values Of The Trigonometric Functions
Problem 38E
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26. What is the solution for the equation
29. It is
below?
the
inc
ú = (sin r)(sin 2u)
Ce
G
be
(a)
cot 2u = I - sin r cos x +c
(b)
sec 2u cot 2u = sin r cos r -x+c
(c) 2 cot 2u
sin r cos r -I+c
%3D
(d)
cot 2u = sin 2r- x+ c
%3D
(e) None of the above
Transcribed Image Text:30 26. What is the solution for the equation 29. It is below? the inc ú = (sin r)(sin 2u) Ce G be (a) cot 2u = I - sin r cos x +c (b) sec 2u cot 2u = sin r cos r -x+c (c) 2 cot 2u sin r cos r -I+c %3D (d) cot 2u = sin 2r- x+ c %3D (e) None of the above
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