solve the equation algebraically

Question

Solve the equation algebraically, and then use a calculator to find the values on the interval [0,2⁢π). Round to 3 decimal places.

tan2⁡x-sec⁡x=1

t=

Answer 1

basic solution: x= pi/3 = 60 degrees also 240 degrees

if you meant tan^2(x) -secx =1

tan^2(60) -sec60 = 3 -2 =1

tan^2(240) -sec240 = 3 -2 = 1

also x = pi

tan^2(180) - sec180 = 0^ - (-1) = 1

but if you really meant

tan^2(x) - sec^2(x) = 1, it's an identity true for any value of x

or

use tan^2(x) - sec^2(x) = 1

then

1 - sec^2(x) -sec(x) = 1

sec^2(x) +secx=0

secx=0 or -1

1/cosx =-1

cosx = -1

x= pi

Answer 2

I'm guessing you meant tan²(x)-sec(x)=1. tan(2x)-sec(x)=1 is very complicated or unsolvable.

Using identity

sec²(x)-1-sec(x) = 1

Let u = sec(x)

u²-u-2=0

(u+1)(u-2) = 0

u = -1, 2

sec(x) = -1, x = π

sec(x) = 2, x= π/3, 5π/3

Answer 3

tan²x = sec² - 1

tan²x - sec x = 1

(sec² -1) - sec x = 1

Can you use Algebraic techniques to solve and answer?