How do you solve #16cos^2x-8=0# and find all solutions in the interval #(-2pi,2pi)#?

1 Answer
Binayaka C.
Nov 3, 2016

Solution : #pi/4, (7pi)/4, -pi/4, -(-7pi)/4# and #x=(3pi)/4, (5pi)/4, (-3pi)/4, (-5pi)/4#

Explanation:

#16cos^2x-8=0 or 16cos^2x=8 or cos^2x=1/2 or cosx= +- 1/sqrt2#

In the interval #(-2pi,2pi)#
#cos (pi/4)= 1/sqrt2 ; cos (2pi-pi/4)= 1/sqrt2 ; cos (-pi/4)= 1/sqrt2 ; cos (-2pi+pi/4)= 1/sqrt2 ; :. x=pi/4, (7pi)/4, -pi/4, -(-7pi)/4#. And

#cos (pi -pi/4)= -1/sqrt2 ; cos (pi+pi/4)= -1/sqrt2 ; cos (-pi+pi/4)= -1/sqrt2 ; cos (-pi-pi//4)= -1/sqrt2 ; :. x=(3pi)/4, (5pi)/4, (-3pi)/4, (-5pi)/4#.{Ans]

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