What is the general solution for 4cos^2(x)+1=0 ?

The general solution for 4cos^2(x)+1=0 is No Solution for x\in\mathbb{R}

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Popular >

4cos^2(x)+1=0

Solution

4 cos^{2} (x) + 1 = 0

Solution

No Solution for x \in R

Solution steps

4 cos^{2} (x) + 1 = 0

Solve by substitution

4 cos^{2} (x) + 1 = 0

Let:

cos (x) = u

4 u^{2} + 1 = 0

4 u^{2} + 1 = 0 : u = i \frac{1}{2}, u = - i \frac{1}{2}

4 u^{2} + 1 = 0

Move

1

to the right side

4 u^{2} + 1 = 0

Subtract

1

from both sides

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

Simplify

4 u^{2} = - 1

4 u^{2} = - 1

Divide both sides by

4

4 u^{2} = - 1

Divide both sides by

4

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

Simplify

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

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - \frac{1}{4}, u = - - \frac{1}{4}

Simplify

- \frac{1}{4} : i \frac{1}{2}

- \frac{1}{4}

Apply radical rule:

- a = - 1 a

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

= - 1 \frac{1}{4}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{4}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

\frac{1}{4} = \frac{1}{4}

= i \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= i \frac{1}{2}

Apply rule

1 = 1

= i \frac{1}{2}

Rewrite

i \frac{1}{2}

in standard complex form:

\frac{1}{2} i

i \frac{1}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 i}{2}

Multiply:

1 i = i

= \frac{i}{2}

= \frac{1}{2} i

Simplify

- - \frac{1}{4} : - i \frac{1}{2}

- - \frac{1}{4}

Simplify

- \frac{1}{4} : i \frac{1}{2}

- \frac{1}{4}

Apply radical rule:

- a = - 1 a

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

= - 1 \frac{1}{4}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{4}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

\frac{1}{4} = \frac{1}{4}

= i \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= i \frac{1}{2}

= - i \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2} i

u = i \frac{1}{2}, u = - i \frac{1}{2}

Substitute back

u = cos (x)

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

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

cos (x) = i \frac{1}{2} :

No Solution

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

No Solution

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

No Solution

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

No Solution

Combine all the solutions

No Solution for x \in R

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)+1=0 ?
The general solution for 4cos^2(x)+1=0 is No Solution for x\in\mathbb{R}