trigonometry help please!!?

6 tan x - sec^2 x = 7

6tanx - (1+tan^2x) =7

6tanx -1 -tan^2x =7

tan^2x -6tanx +8 = 0

(tanx-2)(tanx-4) = 0

tanx=2 ; tanx = 4

key angle

63.4 ; 76.0

tan positive is found in the 1st & 3rd qudrant

for tanx =2

x = 63.4, (180+63.4)=243.4

for tanx =4

x= 76, (180+76) = 256

x=63.4, 76, 243.4, 256

here i'm assuming x is between 0 & 360

You need to give us a little more on the range/ domain of your problem

The quadratic becomes :

tan(x) ^2 -6tan x + 8

so tan x = 4 or tan x = 2

this gives x = about 76° and 63.4° ( appriox )

now do you want it in degrees ? Radians ?? first quadrant only ? all values from 0 to 360° ? The general formula ?

add some detail, and one of us can finish this problem.

replace sec^2 x by 1 + tan^2 x

and put tanx = t

solve the quadratic for t

then put tanx equal to those solutions of quadratic equation

get the desired value of x