2(cos x)^2+2cos x =3(sinx)^2

x=?

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To determine x, we'll re-write the equation so that to caontain only the function cosine.

We'll apply the fundamental formula of trigonometry:

(sinx)^2 + (cos x)^2 = 1

(sinx)^2 = 1-(cos x)^2

We'll re-write the equation:

2(cos x)^2+2cos x = 3(1-(cos x)^2)

We'll remove the brackets:

2(cos x)^2+2cos x = 3 - 3(cos x)^2

We'll move all terms to one side:

2(cos x)^2 + 3(cos x)^2 + 2cos x - 3 = 0

We'll combine like terms:

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

We'll substitute cosx = t

5t^2 + 2t - 3 = 0

We'll apply the qaudratic formula:

t1 = [-2+sqrt(4+60)]/10

t1 = (-2+8)/10

t1 = 6/10

t1 = 3/5

t2 = (-2-8)/10

t2 = -1

cos x = 3/5

**x = +/-arccos3/5 + 2*k*pi**

cos x = -1

x = pi + 2*k*pi

**x = pi*(2*k + 1)**

2cos^2x +2cosx = 3(sinx)^2. To find x.

Solution:

We express the equation in terms of cosx = c say:

2c^2+2c = 3(1-c^2) as (sinx)^2 = 1-cos^2x =1- c^2

2c^2+2c -3(1-c^2) = 0

2c^2+3c^2 +2c -3 = 0

5c^2 +2c -3 = 0

(5c-3 ) (c +1) = 0

c= 2/5 or c =-1

So c= cosx = 2/5 Or x = arc cos (2/5) =

c =-1 gives cos x =-1 . Or x = pi

x = 2npi +or-arccos(2/5), n =0.1,2...

x = (2n+1)pi, n = 0,1,2,...

We have to use this equation: 2 (cos x)^2+ 2cos x= 3 (sin x)^2 to find x.

Here we will use the relation that (cos x)^2+ (sin x)^2=1

So from the equation:

2(cos x)^2+ 2cos x= 3 (sin x)^2 replace (sin x)^2 with 1- (cos x)^2

We get 2(cos x)^2+ 2cos x= 3[ 1- (cos x)^2]

=> 2(cos x)^2+ 2cos x= 3- 3(cos x)^2

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

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

=>5 cosx(cos x+1)-3(cosx +1)=0

=>(5cos x-3)(cosx+1)=0

So either (5cos x-3)=0, cos x=3/5

or (cosx+1)=0, cosx=-1

x=cos inverse (-1)=pi= 3.1415

x=cos inverse (3/5)= 0.9272

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