What is x if 5sin^2x + 3sinxcosx + cos^2x = 3 ?

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neela | High School Teacher | (Level 3) Valedictorian

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5sin^2x +3sinxcosx +cos^2x = 3.

We  know that sin^2x = 1-cos^2x.

Therefore 5(1-cos^2x) +3sinx cosx + cos^2x = 3.

3sinx cosx = 5cos^2x- cos^2x +3-5 =  4cos^2x -2

3sinx cosx = 4cos^2x-2.

We Square both sides:

9 sin^2x cos^2x = (4cos^2x -2)^2.

9(1-c^2) c^2 = (4c^2-2)^2, where c = cosx.

9(c^2-c^4) = 16c^4 -16c^2 +4.

0 = (16+9)c^4  -(16+9)c^2 + 4 .

25c^4 - 25c^2 +4 = 0.

 25c^2 - 20c^2 - 5c^2 +4 = 0.

5c^2(5c^2 -4)  -1(5c^2-4) = 0.

(5c^2 -4)(5c^2-1) = 0.

5c^2 - 4 gives: c^2 = 4/5 Or c = cosx = sqrt(4/5) or cosx = sqrt(4/5).

Therefore cosx = 26.565 deg approx . Or  x = -26.565 deg = 333.435 deg aprox.

Similarely  c = -sqrt(4/5) gives: cosx =  153.435 deg or  206 .56 deg

5c^2-1 = 0 gives:  c = sqrt(1/5) or c = sqrt(1/5) : So cosx = 63.435 degdeg Or  - 63.435 deg = 296.565 deg

c  = - sqrt(1/ 5) gives cosx = -1/srt(1/5) , or x =  116.565 deg or x =  243 .345 deg.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To determine x means to find the angle x from the given identity. We'll transform the given identity into a homogenous equation by substituting 3 by 3*1 = 3[(sin x)^2 + (cos x)^2] and moving all terms to one side.

5(sin x)^2 + 3sinxcosx + (cos x)^2 -3(sin x)^2 - 3(cos x)^2 = 0

We'll combine like terms:

2(sin x)^2 + 3sinx*cosx - 2(cos x)^2 = 0

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

2(sin x)^2/(cos x)^2 + 3sinx*cosx/(cos x)^2 - 2 = 0

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

 2(tan x)^2 + 3tan x - 2 = 0

We'll substitute tan x = t:

2t^2 + 3t - 2 = 0

We'll apply the quadratic formula:

t1 = [-3+sqrt(9+16)]/2

t1 = (-3+5)/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

tan x = t2

tan x = -4

x = - arctan (4) + k*pi

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

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

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