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

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

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

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

Solution

cos (2 x) + 3 cos (x) + 2 = 0, \frac{π}{2} \leq x \leq π

Solution

x = \frac{2 π}{3}, x = π

Degrees

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

Solution steps

cos (2 x) + 3 cos (x) + 2 = 0, \frac{π}{2} \leq x \leq π

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}, \frac{π}{2} \leq x \leq π : x = \frac{2 π}{3}

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

\frac{π}{2} \leq x \leq π

x = \frac{2 π}{3}

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

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

\frac{π}{2} \leq x \leq π

x = π

Combine all the solutions

x = \frac{2 π}{3}, x = π

Popular Examples

2cos(a)=sqrt(3)

2 cos (a) = 3

cos^2(x)-sin(-x)=0

cos^{2} (x) - sin (- x) = 0

sin^2(4x)=1

sin^{2} (4 x) = 1

sinh(x)= 2/3

sinh (x) = \frac{2}{3}

cos(α_{2})= 1/(sqrt(14)),0<= α_{2}<= 2pi

cos (α_{2}) = \frac{1}{14}, 0 \leq α_{2} \leq 2 π

Frequently Asked Questions (FAQ)

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