Solve the equation 5cos^2x-6tanx*cosx=cos^2x-5sin^2x-1.

For the beginning, we'll move all terms to one
side:

5(cos x)^2 - 6tanx*cosx - (cos x)^2 + 5(sin x)^2 + 1
= 0

We'll re-group the
terms:

[1 - (cos x)^2] - 6tanx*cosx + 5[(sin x)^2 + (cos
x)^2] = 0

We'll write the fundamental formula of
trigonometry;

[(sin x)^2 + (cos x)^2] =
1

(sin x)^2 = 1 - (cos x)^2

We
also know that tan x = sin x/cos x

The equation will
become:

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

We'll simplify and we'll
get:

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

We'll substitute sin x =
t

t^2 - 6t + 5 = 0

We'll apply
quadratic formula:

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

t1 = (6+4)/2

t1 =
5

t2 = (6-4)/2

t2 =
1

But sin x = t

sin x =
t1

sin x = 5, impossible since sin x =<
1.

sin x = t2

sin x =
1

x = k*arcsin 1 + kpi

x =
k*pi/2 + kpi

The solution of the equation is:
{k*pi/2 + kpi}.