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

The general solution for cos^2(x)+4cos(x)+4=0 is No Solution for x\in\mathbb{R}

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

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

No Solution for x \in R

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 4, c = 4

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

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

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

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

Multiply the numbers:

4 \cdot 1 \cdot 4 = 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 1}

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

u = - 2

The solution to the quadratic equation is:

u = - 2

Substitute back

u = cos (x)

cos (x) = - 2

cos (x) = - 2

cos (x) = - 2 :

No Solution

cos (x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

No Solution for x \in R

2 cos (x) = tan (x)

sin (x) + sin^{2} (x) = 1

sin^{3} (a) = 3 sin (a) - 4 sin^{3} (a)

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

sin (θ) + 2 sin (θ) cos (θ) = 0

What is the general solution for cos^2(x)+4cos(x)+4=0 ?
The general solution for cos^2(x)+4cos(x)+4=0 is No Solution for x\in\mathbb{R}