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

Q: What is the general solution for 2+cos(2x)=-3cos(x),0<= x<= 2pi ?

The general solution for 2+cos(2x)=-3cos(x),0<= x<= 2pi is x=(2pi)/3 ,x=(4pi)/3 ,x=pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2+cos(2x)=-3cos(x),0<= x<= 2pi

Solution

2 + cos (2 x) = - 3 cos (x), 0 \leq x \leq 2 π

Solution

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

Degrees

x = 12 0^{\circ}, x = 24 0^{\circ}, x = 18 0^{\circ}

Solution steps

2 + cos (2 x) = - 3 cos (x), 0 \leq x \leq 2 π

Subtract

- 3 cos (x)

from both sides

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

Rewrite using trig identities

2 + cos (2 x) + 3 cos (x)

Use the Double Angle identity:

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

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

Simplify

2 + 2 cos^{2} (x) - 1 + 3 cos (x) : 2 cos^{2} (x) + 3 cos (x) + 1

2 + 2 cos^{2} (x) - 1 + 3 cos (x)

Group like terms

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

Add/Subtract the numbers:

2 - 1 = 1

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

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

1 + 2 cos^{2} (x) + 3 cos (x) = 0

Solve by substitution

1 + 2 cos^{2} (x) + 3 cos (x) = 0

Let:

cos (x) = u

1 + 2 u^{2} + 3 u = 0

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

1 + 2 u^{2} + 3 u = 0

Write in the standard form

a x^{2} + b x + c = 0

2 u^{2} + 3 u + 1 = 0

Solve with the quadratic formula

2 u^{2} + 3 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = 3, c = 1

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

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

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

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 3 + 1 = - 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}

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

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

Subtract the numbers:

- 3 - 1 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{2 π}{3}, x = \frac{4 π}{3}

cos (x) = - \frac{1}{2}, 0 \leq x \leq 2 π

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{2 π}{3}, x = \frac{4 π}{3}

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

cos (x) = - 1, 0 \leq x \leq 2 π

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 = π + 2 πn

x = π + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = π

Combine all the solutions

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

Popular Examples

sec(x+10)=5.759

sec (x + 1 0^{\circ}) = 5.759

cos(3x+pi/4)=(sqrt(3))/2

cos (3 x + \frac{π}{4}) = \frac{3}{2}

-sin(4x)=sin(8x)

- sin (4 x) = sin (8 x)

tan(θ)=(3sqrt(3))/3

tan (θ) = \frac{3 3}{3}

cos(2x)=sqrt(1-sin^2(2x)),0<= x<= 2pi

cos (2 x) = 1 - sin^{2} (2 x), 0 \leq x \leq 2 π

Frequently Asked Questions (FAQ)

What is the general solution for 2+cos(2x)=-3cos(x),0<= x<= 2pi ?
The general solution for 2+cos(2x)=-3cos(x),0<= x<= 2pi is x=(2pi)/3 ,x=(4pi)/3 ,x=pi