Find the angle x for sin^2x+sinx*cosx-4cos^2x+1=0

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve sin^2x+sinxcosx-4cos^2+1 = 0.

We know that sin^2x +cos^2x = 1.

So cos^2x = 1-sin^2x. We substitute cos^2x = 1-sin^2x in the given equation:

sin^2x+sinxcosx-4(1-sin^2x) +1 = 0.

sin^2x+sinxcosx-4+4sin^2x +1 = 0.

5sin^2 -3 = -sinxcosx.

squaring, we get:

25sin^4x-30sin^2 +9 = sin^2xcos^2x.

25sin^4x -30sin^x+9= sin^2x(1-sin^2x).

25sin^2x-30sin^2x+9= sin^2x-sin^4x.

26sin^4x - 31sin^2x+9 = 0.

26sin^4x-18sin^2x-13sin^2x +9 = 0.

2sin^2x(13sin^2x-9)-1(13sin^2-9) = 0.

(2sin^2x-1) (13sin^2x+9) = 0.

 2sin^2x-1 = 0 , Or 13sin^2x+9 = 0.

2sin^2 = 1.

sin^2x = 1/2.

sinx = sqrt(1/2) . Or sinx = -sqrt(1/2).

Sinx = 1/2 goves: x = 45 degree or x= 180-45 = 135 degree.

sinx = -1/2 gives x = - 45 degree (= 315deg) Or x = 225 degree.

 13 sin^2 x +9  = 0 does not give any real solution.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To find the angle x means to solve the given equation. We'll transform the given equation into a homogenous equation by substituting 1 by (sin x)^2 + (cos x)^2 = 1.

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

We'll combine like terms:

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

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

2 (tanx)^2 + tan x - 3 = 0

We'll substitute tan x = t:

2t^2 + t - 3 = 0

We'll apply the quadratic formula:

t1 = [-1+sqrt(1+24)]/4

t1 = (-1+5)/4

t1 = 1

t2 = (-1-5)/4

t2 = -3/2

We'll put tan x = t1:

tan x = 1

x = arctan 1 + k*pi

x = pi/4 + k*pi

tan x = t2

tan x = -3/2

x = - arctan (3/2) + k*pi

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

{pi/4 + k*pi} U {- arctan (3/2) + k*pi}

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