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

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

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Popular >

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

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Solve by substitution

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

Let:

cos (4 x) = u

2 u^{2} + 1 = 0

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

2 u^{2} + 1 = 0

Move

1

to the right side

2 u^{2} + 1 = 0

Subtract

1

from both sides

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

Simplify

2 u^{2} = - 1

2 u^{2} = - 1

Divide both sides by

2

2 u^{2} = - 1

Divide both sides by

2

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

Simplify

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

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

For

x^{2} = f (a)

the solutions are

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

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

Simplify

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

- \frac{1}{2}

Apply radical rule:

- a = - 1 a

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

= - 1 \frac{1}{2}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{2}

Simplify

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

- - \frac{1}{2}

Simplify

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

- \frac{1}{2}

Apply radical rule:

- a = - 1 a

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

= - 1 \frac{1}{2}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{2}

= - i \frac{1}{2}

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

Substitute back

u = cos (4 x)

cos (4 x) = i \frac{1}{2}, cos (4 x) = - i \frac{1}{2}

cos (4 x) = i \frac{1}{2}, cos (4 x) = - i \frac{1}{2}

cos (4 x) = i \frac{1}{2} :

No Solution

cos (4 x) = i \frac{1}{2}

No Solution

cos (4 x) = - i \frac{1}{2} :

No Solution

cos (4 x) = - i \frac{1}{2}

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

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