Question #a9636

Question

1717 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-14T11:47:57

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

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

$=> 4x= 2npi$

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

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
$4x = 2kpi$
$x = (2kpi)/4 = (kpi)/2$