sin^2 x - 3sin2x + 5cos^2 x = 0

First let us rewrite:

We know that:

sin2x = 2sinx*cosx

Now substitute:

sin^2 x - 3*2sinx*cosx + 5cos^2 x = 0

sin^2 x - 6sinx*cosx + 5cos^2 x= 0

Nw let us factor:

(sin-5cosx)(sinx - cosx) = 0

Then we have two cases:

sinx -5cosx = 0

==> sinx= 5cosx

==> tanx = 5

**==> x= arctan 5 + 2kpi.**

Also,

sinx-cosx = 0

==> sinx = cosx

**==> x= pi/4 + 2kpi **

The given equation is a homogeneous equation in sin x and cos x and we'll divide the equation by (cos x)^2.

Before dividing by (cos x)^2, we'll write the formula for sin 2x:

sin 2x = sin (x+x) = sin x*cos x + sinx*cos x

sin 2x = 2 sin x*cos x

We'll substitute sin 2x by it's formula in the given equation:

(sin x)^2 - 3*2*sin x*cos x + 5 (cos x)^2 = 0

Now, we can divide by (cos x)^2:

(sin x/cos x)^2 - 6(sinx/cosx) + 5 = 0

But the ratio sin x/cos x = tan x

We'll substitute the ratio by the function tan x:

(tan x)^2 - 6tan x + 5 = 0

We'll substitute tan x = t

t^2 - 6t + 5 = 0

We'll apply the quadratic formula:

t1 = [6+sqrt(36 - 20)]/2

t1 = (6+4)/2

t1 = 5

t2 = 1

tan x = t1

tan x = 5

**x = arctan 5 + k*pi**

tan x = t2

tan x = 1

x = arctan 1 + k*pi

**x = pi/4 + k*pi**

**The solution of the equation are: {arctan 5 + k*pi}U{pi/4 + k*pi}.**