What is the general solution for 2cos^2(x)+cos(x)-3=0 ?

The general solution for 2cos^2(x)+cos(x)-3=0 is x=2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

2 u^{2} + u - 3 = 0

2 u^{2} + u - 3 = 0 : u = 1, u = - \frac{3}{2}

2 u^{2} + u - 3 = 0

Solve with the quadratic formula

2 u^{2} + u - 3 = 0

Quadratic Equation Formula:

For

a = 2, b = 1, c = - 3

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 3) = 5

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 3)

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 \cdot 2}, u_{2} = \frac{- 1 - 5}{2 \cdot 2}

u = \frac{- 1 + 5}{2 \cdot 2} : 1

\frac{- 1 + 5}{2 \cdot 2}

Add/Subtract the numbers:

- 1 + 5 = 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 - 5}{2 \cdot 2} : - \frac{3}{2}

\frac{- 1 - 5}{2 \cdot 2}

Subtract the numbers:

- 1 - 5 = - 6

= \frac{- 6}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 6}{4}

Apply the fraction rule:

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

= - \frac{6}{4}

Cancel the common factor:

2

= - \frac{3}{2}

The solutions to the quadratic equation are:

u = 1, u = - \frac{3}{2}

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = - \frac{3}{2}

cos (x) = 1, cos (x) = - \frac{3}{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{3}{2} :

No Solution

cos (x) = - \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = 2 πn

Popular Examples

5sin(θ)tan(θ)-10tan(θ)+3sin(θ)-6=0

5 sin (θ) tan (θ) - 10 tan (θ) + 3 sin (θ) - 6 = 0

3cos^2(θ)+6cos(θ)-4=0

3 cos^{2} (θ) + 6 cos (θ) - 4 = 0

arccos(x)=1

arccos (x) = 1

cos(x)=sqrt(3/4)

cos (x) = \frac{3}{4}

sin(x)=sin(x/2)

sin (x) = sin (\frac{x}{2})

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)+cos(x)-3=0 ?
The general solution for 2cos^2(x)+cos(x)-3=0 is x=2pin