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

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### 2 Answers

2sin^2x+5sinxcosx +5cos^2x=1.

5sinxcosx = 1-5cos^2x-2sin^2x.

5sinxcosx = 1-5(1-sin^2x)-2sin^2x.

5sinxcosx = 1-5+5sin^2x-2sin^2x

5sinxcosx = 3sin^2-4.

We square both sides:

25sin^2xcos^2x = (3sin^2x-4)^2

25sin^2x (1-sin^2x) = 9 sin^4x-24sin^2x+16

25sin^2-25sin^4 = 9sin^4 -24sin^2x+16.

9s^4+25s^4 - 49s^2 +16 = 0, where s= sinx.

34s^4 -49s^2+16 = 0.

34s^4 - 17s^2 - 32s^2+16 = 0

17s^2(2s^2-1)-16(2s^2-1) = 0

(2s^2-1((17s^2-16) = 0

s^2= 1/2, Or s^2 = 16/17

s= +sqrt1/2 , sinx = sqrt(1/2), or x = 45 or 135 deg

s= -sqrt(1/2)., sinx = -sqrt(1/2), or x= 225, or x= 360-45 = 315 deg

s = sqrt(16/17) , sinx = sqr(16/17) , x = 75.96 deg, or 104.04 deg

s = -sqrt(16/17), sinx = -sqr(16/17), x = -75.96 deg = 360-75.96 = 284.94 deg or 180+75.96 = 255.96 deg

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}.**