Question

How do you solve `2tan^2x - 3secx = 0`?

Answer 1

If we restrict the domain of `x` to `[0, 2pi)`
then `x=(pi)/3 or (5pi)/3`

Explanation:

`2tan^2(x) - 3sec(x) = 0`

`rarr 2(sin^2(x))/(cos^2(x)) - 3(1/cos(x)) = 0`

`color(white)("XXXX")`assuming `cos(x)!=0` (the equation would not be valid if it were)
`color(white)("XXXX")`Multiply by `cos^2(x)`
`2sin^2(x) = 3cos(x)`
`color(white)("XXXX")`Use Pythagorean relationship
`rarr 2(1-cos^2(x)) -3cos(x) = 0`
`color(white)("XXXX")`Simplify
`rarr 2cos^2(x)+3cos(x)-2=0`
`color(white)("XXXX")`Factor
`rarr (2cos(x)-1)(cos(x)+2) = 0`

`rArr 2cos(x)-1 = 0`
or
`color(white)("XXXX")cos(x)+2 = 0`

*Case 1
If `2cos(x)-1 = 0`
`color(white)("XXXX")rarr cos(x) = 1/2`
`color(white)("XXXX")rarr x = (pi)/3 or (5pi)/3`
`color(white)("XXXX")color(white)("XXXX")`assuming `x in [0,2pi)`

Case 2
If `cos(x)+2 = 0`
`color(white)("XXXX")rarr cos(x) = -2`
BUT
`color(white)("XXXX")cos(x) in [-1,+1] " for all " x`
Therefore this case is extraneous.