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

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}.