What is the general solution for sin^2(θ)-2sin(θ)+3=0 ?

The general solution for sin^2(θ)-2sin(θ)+3=0 is No Solution for θ\in\mathbb{R}

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Popular >

sin^2(θ)-2sin(θ)+3=0

Solution

sin^{2} (θ) - 2 sin (θ) + 3 = 0

Solution

No Solution for θ \in R

Solution steps

sin^{2} (θ) - 2 sin (θ) + 3 = 0

Solve by substitution

sin^{2} (θ) - 2 sin (θ) + 3 = 0

Let:

sin (θ) = u

u^{2} - 2 u + 3 = 0

u^{2} - 2 u + 3 = 0 : u = 1 + 2 i, u = 1 - 2 i

u^{2} - 2 u + 3 = 0

Solve with the quadratic formula

u^{2} - 2 u + 3 = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = 3

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 3 : 22 i

(- 2)^{2} - 4 \cdot 1 \cdot 3

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

= 2^{2} - 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 2^{2} - 12

Apply imaginary number rule:

- a = i a

= i 12 - 2^{2}

- 2^{2} + 12 = 22

- 2^{2} + 12

2^{2} = 4

= - 4 + 12

Add/Subtract the numbers:

- 4 + 12 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

= 22 i

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

2 + 22 i : 2 (1 + 2 i)

2 + 22 i

Rewrite as

= 2 \cdot 1 + 22 i

Factor out common term

2

= 2 (1 + 2 i)

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

Divide the numbers:

\frac{2}{2} = 1

= 1 + 2 i

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

2 - 22 i : 2 (1 - 2 i)

2 - 22 i

Rewrite as

= 2 \cdot 1 - 22 i

Factor out common term

2

= 2 (1 - 2 i)

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

Divide the numbers:

\frac{2}{2} = 1

= 1 - 2 i

The solutions to the quadratic equation are:

u = 1 + 2 i, u = 1 - 2 i

Substitute back

u = sin (θ)

sin (θ) = 1 + 2 i, sin (θ) = 1 - 2 i

sin (θ) = 1 + 2 i, sin (θ) = 1 - 2 i

sin (θ) = 1 + 2 i :

No Solution

sin (θ) = 1 + 2 i

No Solution

sin (θ) = 1 - 2 i :

No Solution

sin (θ) = 1 - 2 i

No Solution

Combine all the solutions

No Solution for θ \in R

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(θ)-2sin(θ)+3=0 ?
The general solution for sin^2(θ)-2sin(θ)+3=0 is No Solution for θ\in\mathbb{R}