Question #a9636

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Answer 1 · 2016-12-14T11:47:57

#cos ( 2x + 1 )= cos( 2x - 1)#

#=>( 2x +1)=2npi-( 2x -1)," where "n in ZZ#

#=> 4x= 2npi#

#=> x= 1/2(npi)#

Answer 2 · 2016-12-14T15:37:56

#x = (kpi)/2#

Reminder:
cos x = cos a --> #x = +- a + 2kpi#
Note: the equation can be written in radian form -->
#cos (2x + pi/3.14) = cos (2x - pi/3.14)#

Solve:
cos (2x + 1) = cos (2x - 1)
#2x+ 1 = +- (2x - 1) + 2kpi#

a. 2x + 1 = 2x - 1 + 2kpi
Equation insolvable

b. 2x + 1 = - 2x + 1 + 2kpi
#4x = 2kpi#
#x = (2kpi)/4 = (kpi)/2#