Question

2sec^2x=5-tanx

Answer 1

do you want to solve for x?

Answer 2

sec(x) = 1/cos(x)
tan(x) = sin(x)/cos(x)

Answer 3

use sin^2 + cos^2 = 1

divide each term by cos^1

so sin^2/cos^2 + cos^2/cos^2 = 1/Cos^2

or tan^2 + 1 = sec^2

use this substitution

then 2(tan^2(x) + 1) = 5 - tan(x)

rewriting

2tan^2(x) + tan(x) - 3 = 0

solve the quadratic by factorising

(2tan(x) +3)(tan(x) - 1) = 0

then tan(x) = -3/2 and 1

now find all the solutions tan is positive in 1st and 3rd quadrants negative in 2nd and 4th

\[\tan^{-1} (1) = 45\]

x = 45 and 225

\[\tan^{-1}3/2 = 56.3\]

2nd quadrant 180 - 56.3 4th quadrant 360 - 56.3

they are all the solutions over the domain 0<= x <= 360