Thursday, June 4, 2015

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

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}.

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