What is the general solution for 2cos^2(x)+3cos(x)=4cos(x)+1 ?

The general solution for 2cos^2(x)+3cos(x)=4cos(x)+1 is x=2pin,x=(2pi)/3+2pin,x=(4pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos^2(x)+3cos(x)=4cos(x)+1

Solution

2 cos^{2} (x) + 3 cos (x) = 4 cos (x) + 1

Solution

x = 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{2} (x) + 3 cos (x) = 4 cos (x) + 1

Solve by substitution

2 cos^{2} (x) + 3 cos (x) = 4 cos (x) + 1

Let:

cos (x) = u

2 u^{2} + 3 u = 4 u + 1

2 u^{2} + 3 u = 4 u + 1 : u = 1, u = - \frac{1}{2}

2 u^{2} + 3 u = 4 u + 1

Move

1

to the left side

2 u^{2} + 3 u = 4 u + 1

Subtract

1

from both sides

2 u^{2} + 3 u - 1 = 4 u + 1 - 1

Simplify

2 u^{2} + 3 u - 1 = 4 u

2 u^{2} + 3 u - 1 = 4 u

Move

4 u

to the left side

2 u^{2} + 3 u - 1 = 4 u

Subtract

4 u

from both sides

2 u^{2} + 3 u - 1 - 4 u = 4 u - 4 u

Simplify

2 u^{2} - u - 1 = 0

2 u^{2} - u - 1 = 0

Solve with the quadratic formula

2 u^{2} - u - 1 = 0

Quadratic Equation Formula:

For

a = 2, b = - 1, c = - 1

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

(- 1)^{2} - 4 \cdot 2 (- 1) = 3

(- 1)^{2} - 4 \cdot 2 (- 1)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 3}{2 \cdot 2}, u_{2} = \frac{- ( - 1 ) - 3}{2 \cdot 2}

u = \frac{- ( - 1 ) + 3}{2 \cdot 2} : 1

\frac{- ( - 1 ) + 3}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 + 3}{2 \cdot 2}

Add the numbers:

1 + 3 = 4

= \frac{4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 1 ) - 3}{2 \cdot 2} : - \frac{1}{2}

\frac{- ( - 1 ) - 3}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 - 3}{2 \cdot 2}

Subtract the numbers:

1 - 3 = - 2

= \frac{- 2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = 1, u = - \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = - \frac{1}{2}

cos (x) = 1, cos (x) = - \frac{1}{2}

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = - \frac{1}{2}

General solutions for

cos (x) = - \frac{1}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Combine all the solutions

x = 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Popular Examples

6sin^2(x)+5sin(x)=0

6 sin^{2} (x) + 5 sin (x) = 0

sin(x)=-1/2 ,(0,2pi)

sin (x) = - \frac{1}{2}, (0, 2 π)

tan(θ)= 4/3 ,cos(θ)<0

tan (θ) = \frac{4}{3}, cos (θ) < 0

cot(x)=(sin(x))/(cos(x))

cot (x) = \frac{sin ( x )}{cos ( x )}

0.6=(cosh(0.2m))/(cosh(0.4m))

0.6 = \frac{cosh ( 0.2 m )}{cosh ( 0.4 m )}

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)+3cos(x)=4cos(x)+1 ?
The general solution for 2cos^2(x)+3cos(x)=4cos(x)+1 is x=2pin,x=(2pi)/3+2pin,x=(4pi)/3+2pin