What is the general solution for cos(2x)+4cos(x)+3=0 ?

The general solution for cos(2x)+4cos(x)+3=0 is x=pi+2pin

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cos(2x)+4cos(x)+3=0

Solution

cos (2 x) + 4 cos (x) + 3 = 0

Solution

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (2 x) + 4 cos (x) + 3 = 0

Rewrite using trig identities

3 + cos (2 x) + 4 cos (x)

Use the Double Angle identity:

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

3 - 1 = 2

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

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

2 + 2 cos^{2} (x) + 4 cos (x) = 0

Solve by substitution

2 + 2 cos^{2} (x) + 4 cos (x) = 0

Let:

cos (x) = u

2 + 2 u^{2} + 4 u = 0

2 + 2 u^{2} + 4 u = 0 : u = - 1

2 + 2 u^{2} + 4 u = 0

Write in the standard form

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

2 u^{2} + 4 u + 2 = 0

Solve with the quadratic formula

2 u^{2} + 4 u + 2 = 0

Quadratic Equation Formula:

For

a = 2, b = 4, c = 2

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

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

4^{2} - 4 \cdot 2 \cdot 2 = 0

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

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- 4 \pm 0}{2 \cdot 2}

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

\frac{- 4}{2 \cdot 2} = - 1

\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

u = - 1

The solution to the quadratic equation is:

u = - 1

Substitute back

u = cos (x)

cos (x) = - 1

cos (x) = - 1

cos (x) = - 1 : x = π + 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 = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)+4cos(x)+3=0 ?
The general solution for cos(2x)+4cos(x)+3=0 is x=pi+2pin