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

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

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

Solution

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

Solution

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

Degrees

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

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Move

1

to the left side

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

Add

1

to both sides

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

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = 4, c = 1

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

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

4^{2} - 4 \cdot 4 \cdot 1 = 0

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 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 4}

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

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

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{8}

Apply the fraction rule:

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

= - \frac{4}{8}

Cancel the common factor:

4

= - \frac{1}{2}

u = - \frac{1}{2}

The solution to the quadratic equation is:

u = - \frac{1}{2}

Substitute back

u = cos (x)

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

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

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

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tan (x) = 6

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csc(θ)+1.597=0

csc (θ) + 1.597 = 0

Frequently Asked Questions (FAQ)

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