Question

How do you solve `sec^2x-1+tanx-sqrt3tanx=sqrt3` for `0<=x<=2pi`?

Answer 1

`x=(3pi)/4`, `(7pi)/4`, `(pi)/3`, `(4pi)/3`

Explanation:

`sec^2x-1+tanx-sqrt(3)tanx=sqrt3` for`0 <= x <= 2pi`

Using the Pythagorean identity `tan^2x+1=sec^2x`, substitute for the `sec^2x` term.

`(tan^2x+1)-1+tanx-sqrt(3)tanx=sqrt3`

`tan^2x+tanx-sqrt(3)tanx=sqrt3color(white)(11aaa11)`Combine like terms

`tan^2x+tanx-sqrt3tanx-sqrt3=0color(white)(aaa)`Subtract `sqrt3` from both `color(white)(AaaaaaaaaaaaaaaaaaaaaaaAAAAAAAAAAA)`sides

`tan^2x +(1-sqrt3)tanx-sqrt3=0color(white)(aaaaa)`Group the 2 tan terms

`(tanx+1)(tanx-sqrt3)=0color(white)(aaaaaaaaa)`Factor

`tanx=-1color(white)(aaaaa)tanx=sqrt3color(white)(aaaaa)`Set factors = 0 and solve

`x=(3pi)/4`,`(7pi)/4color(white)(aaaaaa)x=(pi)/3`,`(4pi)/3color(white)(aaaa)`Use the unit circle*

*To find values for `tanx=-1` on the unit circle, `sinx` and `cosx` must be equal but with opposite signs. These conditions are found at `(3pi)/4` where `sinx=sqrt2/2`, `cosx=-sqrt2/2` and `tanx=sinx/cosx=-1`
and at `(7pi)/4` where `sinx=-sqrt2/2`, `cosx=sqrt2/2` and
`tanx=sinx/cosx=-1`

Similarly, to find values for `tanx=sqrt3` on the unit circle,
`tanx=sinx/cosx=sqrt3/2-:1/2=sqrt3` at `pi/3` and
`tanx=sinx/cosx=-sqrt3/2-:-1/2=sqrt3` at `(4pi)/3`