What is the general solution for 4cos^2(x)+8cos(x)+4=0 ?

The general solution for 4cos^2(x)+8cos(x)+4=0 is x=pi+2pin

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4cos^2(x)+8cos(x)+4=0

4 cos^{2} (x) + 8 cos (x) + 4 = 0

x = π + 2 πn

Degrees

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

Solution steps

4 cos^{2} (x) + 8 cos (x) + 4 = 0

Solve by substitution

4 cos^{2} (x) + 8 cos (x) + 4 = 0

Let:

cos (x) = u

4 u^{2} + 8 u + 4 = 0

4 u^{2} + 8 u + 4 = 0 : u = - 1

4 u^{2} + 8 u + 4 = 0

Solve with the quadratic formula

4 u^{2} + 8 u + 4 = 0

Quadratic Equation Formula:

For

a = 4, b = 8, c = 4

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 4 \cdot 4}{2 \cdot 4}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 4 \cdot 4}{2 \cdot 4}

8^{2} - 4 \cdot 4 \cdot 4 = 0

8^{2} - 4 \cdot 4 \cdot 4

Multiply the numbers:

4 \cdot 4 \cdot 4 = 64

= 8^{2} - 64

8^{2} = 64

= 64 - 64

Subtract the numbers:

64 - 64 = 0

= 0

u_{1, 2} = \frac{- 8 \pm 0}{2 \cdot 4}

u = \frac{- 8}{2 \cdot 4}

\frac{- 8}{2 \cdot 4} = - 1

\frac{- 8}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 8}{8}

Apply the fraction rule:

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

= - \frac{8}{8}

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

4 sin (x) cos (x) = 22 cos (x)

t, x = cos^{2} (t)

sin (x) = \frac{24}{27}

cot^{2} (x) = 0

4 cos^{2} (x) + 5 sin (x) = 3

What is the general solution for 4cos^2(x)+8cos(x)+4=0 ?
The general solution for 4cos^2(x)+8cos(x)+4=0 is x=pi+2pin