Question

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

Answer 1

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]