3sec^2 x = 5

First we will rewrite secx = 1/cosx

==> 3 ( 1/cosx)^2 = 5

Now we will divide by 3

==> (1/cos^2 x) = 5/3

==> cos^2 x = 3/5 = 0.6

Now we will take the square root.

==> cosx = +- sqrt(0.6)

==> cosx = +-0.7746

==> x = 39.23

For the interval (0.2pi)

==> x1= 39.23

==> x2= pi - 39.23 = 140.77

==> x3= pi + 39.23 = 219.23

==> x3= 2pi - 39.23 = 320.77

**==> x = { 39.23, 140.77, 219.23, 320.77}**

We have to solve 3*(sec x)^2 = 5 for x between 0 < x < 2*pi

3*(sec x)^2 = 5

3*(1/ cos x)^2 = 5

=> (cos x)^2 = 3/5

cos x = sqrt (3/5) and cos x = -sqrt (3/5)

x = arc cos (sqrt(3/5))

x = 39.23 degrees and 320.768 degrees.

cos x = -sqrt(3/5)

=> x = arc cos(- sqrt (3/5))

x = 140.76 and x = 219.23

**The required values of x are {39.23, 140.76, 219.23, 320.768}**

We know that sec x = 1/cosx

We'll raise to square:

(sec x)^2 = (1/cos x)^2

But (1/cos x)^2 = 1 + (tan x)^2

3(sec x)^2 = 3[1 + (tan x)^2]

The equation will become:

3 + 3(tan x)^2 = 5

3(tan x)^2 = 2

(tan x)^2 = 2/3

tan x = (sqrt6)/3

x = arctan [(sqrt6)/3] + kpi

tan x = -(sqrt6)/3

x = -arctan [(sqrt6)/3] + kpi

**The solutions of the equation are: {arctan [(sqrt6)/3]} ; {arctan [(sqrt6)/3] + pi} ; {- arctan [(sqrt6)/3]} ; {- arctan [(sqrt6)/3] + pi}.**