Question

How do you solve (2(sin^2)x-1)((tan^2)x-3)=0 over the interval (0,2pi)?

Answer 1

See below

Explanation:

graph{(2(sinx)^2-1)((tanx)^2-3) [-10, 10, -5, 5]}

Set the factors equal to 0:

`2sin^2x-1=0`
`tan^2x-3=0`

`sinx= +-sqrt2/2`
`x=pi/4, (3pi)/4, (5pi)/4, (7pi)/4`

`tanx=+-sqrt3`
`x=pi/3, (2pi)/3, (4pi)/3, (5pi)/3`