# Solve the equation sin^2x-sinxcosx-2cos^2x=0.

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

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

But the ratio sin x/cos x = tan x

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

(tan x)^2 - tan x - 2 = 0

We'll substitute tan x = t

t^2 - t - 2 = 0

We'll apply the quadratic formula:

t1 = [1 + sqrt(1+8)]/2

t1 = (1+3)/2

t1 = 2

t2 = -1

tan x = t1

tan x = 2

**x = arctan 2 + k*pi**

tan x = t2

tan x = -1

x = arctan(-1) + k*pi

x = - arctan 1 + k*pi

x = -pi/4 + + k*pi

**x = 3pi/4 + k*pi**

**The solutions of the equation are: {arctan 2 + k*pi}U{3pi/4 + k*pi}.**