How do you solve #2sec2x - cot2x = tan2x# from 0 to 2pi?

2 Answers
Sonnhard
May 26, 2018

#x=pi/12,x=5*pi/12,x=13*pi/12,x=17*pi/12#

Explanation:

Simplifying and factorizing the given equation we obtain
#(-2+csc(2x))sec(2x)=0#

Nghi N.
May 26, 2018

#pi/12; (13pi)/12, and (5pi)/12 ; (17pi)/12#

Explanation:

Call X = 2x, the equation becomes:
#2/(cos X) - cos X/(sin X) - sin X/(cos X) = 0#
#(2sin X - cos^2 x - sin^2 x)/(sin X.cos x) = 0#
#2sin X - 1 + sin^2 X - sin^2 X = 0#
Condition --> #(sin X.cos X) != 0#
#2sin X = 1# --> #sin X = 1/2# --> #sin 2x = 1/2#
Trig table and unit circle give 2 general solutions for 2x -->
#2x = pi/6 + 2kpi# and #2x = (5pi)/6 + 2kpi#
a. #2x = pi/6 + 2kpi# --> #x = pi/12 + kpi#
b. #2x = (5pi)/6 + 2kpi# --> #x = (5pi)/12 + kpi#
From 0 to #2pi#, there are 4 answers (k = 1):
#pi/12#; and #pi/12 + pi = (13pi)/12#
#(5pi)/12#; and #(5pi)/12 + pi = (17pi)/12#

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