# Solve for x2sin^2x+5sinxcosx+5cos^2x=1

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### 1 Answer

To solve for x means to find the angle x from the given identity. We'll transform the given identity into a homogenous equation by substituting 1 by (sin x)^2 + (cos x)^2 = 1 and moving all terms to one side.

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

We'll combine like terms:

(sin x)^2 + 5sinx*cosx + 4(cos x)^2 = 0

Since cos x is different from zero, we'll divide the entire equation by (cos x)^2:

(sin x)^2/(cos x)^2 + 5sinx*cosx/(cos x)^2 + 4 = 0

According to the rule, the ratio sin x/cos x = tan x.

(tan x)^2 + 5tan x + 4 = 0

We'll substitute tan x = t:

t^2 + 5t + 4 = 0

We'll apply the quadratic formula:

t1 = [-5+sqrt(25-16)]/2

t1 = (-5+3)/2

t1 = -1

t2 = (-5-3)/2

t2 = -4

We'll put tan x = t1:

tan x = -1

x = arctan -1 + k*pi

x = -pi/4 + k*pi

-pi/4 = pi - pi/4 = 3pi/4

-pi/4 = 2pi - pi/4 = 7pi/4

x = 3pi/4 + k*pi

x = 7pi/4 + k*pi

tan x = t2

tan x = -4

x = - arctan (4) + k*pi

The solutions of the equation are the values of x angle:

{3pi/4 + k*pi} U {7pi/4 + k*pi} U {- arctan (4) + k*pi}