What is the general solution for sin^2(x)+5sin(x)+6=0 ?

The general solution for sin^2(x)+5sin(x)+6=0 is No Solution for x\in\mathbb{R}

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Popular >

sin^2(x)+5sin(x)+6=0

Solution

sin^{2} (x) + 5 sin (x) + 6 = 0

Solution

No Solution for x \in R

Solution steps

sin^{2} (x) + 5 sin (x) + 6 = 0

Solve by substitution

sin^{2} (x) + 5 sin (x) + 6 = 0

Let:

sin (x) = u

u^{2} + 5 u + 6 = 0

u^{2} + 5 u + 6 = 0 : u = - 2, u = - 3

u^{2} + 5 u + 6 = 0

Solve with the quadratic formula

u^{2} + 5 u + 6 = 0

Quadratic Equation Formula:

For

a = 1, b = 5, c = 6

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot 6 = 1

5^{2} - 4 \cdot 1 \cdot 6

Multiply the numbers:

4 \cdot 1 \cdot 6 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 5 \pm 1}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 5 + 1}{2 \cdot 1}, u_{2} = \frac{- 5 - 1}{2 \cdot 1}

u = \frac{- 5 + 1}{2 \cdot 1} : - 2

\frac{- 5 + 1}{2 \cdot 1}

Add/Subtract the numbers:

- 5 + 1 = - 4

= \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 = \frac{- 5 - 1}{2 \cdot 1} : - 3

\frac{- 5 - 1}{2 \cdot 1}

Subtract the numbers:

- 5 - 1 = - 6

= \frac{- 6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 6}{2}

Apply the fraction rule:

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

= - \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= - 3

The solutions to the quadratic equation are:

u = - 2, u = - 3

Substitute back

u = sin (x)

sin (x) = - 2, sin (x) = - 3

sin (x) = - 2, sin (x) = - 3

sin (x) = - 2 :

No Solution

sin (x) = - 2

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = - 3 :

No Solution

sin (x) = - 3

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

No Solution for x \in R

Popular Examples

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- 6 tan^{2} (θ) + 11 tan (θ) - 6 = - 2 tan (θ)

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

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

12sin^2(x)-cos(x)-6=0

12 sin^{2} (x) - cos (x) - 6 = 0

tan(x/2)-1=0

tan (\frac{x}{2}) - 1 = 0

sin(θ)=-cos(θ)

sin (θ) = - cos (θ)

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)+5sin(x)+6=0 ?
The general solution for sin^2(x)+5sin(x)+6=0 is No Solution for x\in\mathbb{R}