Trying to solve this equation between {0,2Pi): tan(x) + sec(x) = sq.rt3. Tried it several ways to no avail.

tan x = sqrt(3) - sec x
Square both sides
tan^2 x = 3 - 2sqrt(3) sec x + sec^2 x
sec^2 x - 1 = 3 - 2sqrt(3) sec x + sec^2 x
sec x = 2/sqrt(3)
cos x = sqrt(3)/2
x = pi/6, 11pi6(rejected after checking)
Answer: x = pi/6

Sketch the sides of a right triangle with sin x = 1/2
x is 30 degrees or pi/6
tan(x) + sec(x) = 1/sqrt(3) + 2/sqrt(3)
= 3/sqrt(3) = sqrt(3)
(I just tried my luck with a well known triangle)

A better way to go would have been
sin x = s, cos x = c
s/c + 1/c = (s + 1)/c = sqrt(3)
(s + 1)^2 = 3c^2 = 3 - 3s^2
4s^2 + 2s - 2 = 0
2(2s - 1)(s + 1) = 0
s = 1/2 or -1
this would seem to give
x = pi/6 or x = - pi/2
but since cos(- pi/2) = 0
tan(- pi/2) is undefined
the only valid solution is
x = pi/6 ~ 0.5235987...

Regards - Ian

tanx + secx = √3
sinx / cosx + 1 / cosx = √3
sinx + 1 = (√3)cosx
sinx - (√3)cosx = -1

sinx - (√3)cosx = rsin(x + α)
sinx - (√3)cosx = rsinxcosα + rcosxsinα
sinx - (√3)cosx = (kcosα)sinx + (ksinα)cosx

r = √[1² + (-√3)²]
r = √[1² + (-√3)²]
r = √(1 + 3)
r = √4
r = 2

rcosα = 1
rsinα = -√3
rsinα / rcosα = -√3
tanα = -√3
α = 5π / 3

sinx - (√3)cosx = 2sin(x + 5π / 3)
2sin(x + 5π / 3) = -1, [-5π / 3 < x + 5π / 3 < π / 3]
sin(x + 5π / 3) = -1 / 2, [-5π / 3 < x + 5π / 3 < π / 3]
x + 5π / 3 = -5π / 6, -π / 6
x = 5π / 6, 3π / 2

I must admit I have made a mistake somewhere.