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

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

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Popular >

2sin^2(x)+2sin(x)+1=0

Solution

2 sin^{2} (x) + 2 sin (x) + 1 = 0

Solution

No Solution for x \in R

Solution steps

2 sin^{2} (x) + 2 sin (x) + 1 = 0

Solve by substitution

2 sin^{2} (x) + 2 sin (x) + 1 = 0

Let:

sin (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = 2, c = 1

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

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

Simplify

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

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 2^{2} - 8

Apply imaginary number rule:

- a = i a

= i 8 - 2^{2}

- 2^{2} + 8 = 2

- 2^{2} + 8

2^{2} = 4

= - 4 + 8

Add/Subtract the numbers:

- 4 + 8 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= 2 i

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

Separate the solutions

u_{1} = \frac{- 2 + 2 i}{2 \cdot 2}, u_{2} = \frac{- 2 - 2 i}{2 \cdot 2}

u = \frac{- 2 + 2 i}{2 \cdot 2} : - \frac{1}{2} + i \frac{1}{2}

\frac{- 2 + 2 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 + 2 i}{4}

Factor

- 2 + 2 i : 2 (- 1 + i)

- 2 + 2 i

Rewrite as

= - 2 \cdot 1 + 2 i

Factor out common term

2

= 2 (- 1 + i)

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

Cancel the common factor:

2

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

Rewrite

\frac{- 1 + i}{2}

in standard complex form:

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

\frac{- 1 + i}{2}

Apply the fraction rule:

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

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

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

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

u = \frac{- 2 - 2 i}{2 \cdot 2} : - \frac{1}{2} - i \frac{1}{2}

\frac{- 2 - 2 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 - 2 i}{4}

Factor

- 2 - 2 i : - 2 (1 + i)

- 2 - 2 i

Rewrite as

= - 2 \cdot 1 - 2 i

Factor out common term

2

= - 2 (1 + i)

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

Cancel the common factor:

2

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

Rewrite

- \frac{1 + i}{2}

in standard complex form:

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

- \frac{1 + i}{2}

Apply the fraction rule:

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

\frac{1 + i}{2} = - (\frac{1}{2}) - (\frac{i}{2})

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

Remove parentheses:

(a) = a

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

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

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

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

No Solution

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

No Solution

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

No Solution

sin (x) = - \frac{1}{2} - 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 2sin^2(x)+2sin(x)+1=0 ?
The general solution for 2sin^2(x)+2sin(x)+1=0 is No Solution for x\in\mathbb{R}