What is the general solution for 5sin^2(x)+10sin(x)+2=0 ?

The general solution for 5sin^2(x)+10sin(x)+2=0 is x=-0.22735…+2pin,x=pi+0.22735…+2pin

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Popular >

5sin^2(x)+10sin(x)+2=0

Solution

5 sin^{2} (x) + 10 sin (x) + 2 = 0

Solution

x = - 0.22735 \dots + 2 πn, x = π + 0.22735 \dots + 2 πn

Degrees

x = - 13.02659 \dots^{\circ} + 36 0^{\circ} n, x = 193.02659 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 sin^{2} (x) + 10 sin (x) + 2 = 0

Solve by substitution

5 sin^{2} (x) + 10 sin (x) + 2 = 0

Let:

sin (x) = u

5 u^{2} + 10 u + 2 = 0

5 u^{2} + 10 u + 2 = 0 : u = \frac{- 5 + 15}{5}, u = - \frac{5 + 15}{5}

5 u^{2} + 10 u + 2 = 0

Solve with the quadratic formula

5 u^{2} + 10 u + 2 = 0

Quadratic Equation Formula:

For

a = 5, b = 10, c = 2

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 5 \cdot 2}{2 \cdot 5}

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 5 \cdot 2}{2 \cdot 5}

1 0^{2} - 4 \cdot 5 \cdot 2 = 215

1 0^{2} - 4 \cdot 5 \cdot 2

Multiply the numbers:

4 \cdot 5 \cdot 2 = 40

= 1 0^{2} - 40

1 0^{2} = 100

= 100 - 40

Subtract the numbers:

100 - 40 = 60

= 60

Prime factorization of

60 : 2^{2} \cdot 3 \cdot 5

60

60

divides by

260 = 30 \cdot 2

= 2 \cdot 30

30

divides by

230 = 15 \cdot 2

= 2 \cdot 2 \cdot 15

15

divides by

315 = 5 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 5

2, 3, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3 \cdot 5

= 2^{2} \cdot 3 \cdot 5

= 2^{2} \cdot 3 \cdot 5

Apply radical rule:

= 2^{2} 3 \cdot 5

Apply radical rule:

2^{2} = 2

= 2 3 \cdot 5

Refine

= 215

u_{1, 2} = \frac{- 10 \pm 2 15}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- 10 + 2 15}{2 \cdot 5}, u_{2} = \frac{- 10 - 2 15}{2 \cdot 5}

u = \frac{- 10 + 2 15}{2 \cdot 5} : \frac{- 5 + 15}{5}

\frac{- 10 + 2 15}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 10 + 2 15}{10}

Factor

- 10 + 215 : 2 (- 5 + 15)

- 10 + 215

Rewrite as

= - 2 \cdot 5 + 215

Factor out common term

2

= 2 (- 5 + 15)

= \frac{2 ( - 5 + 15 )}{10}

Cancel the common factor:

2

= \frac{- 5 + 15}{5}

u = \frac{- 10 - 2 15}{2 \cdot 5} : - \frac{5 + 15}{5}

\frac{- 10 - 2 15}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 10 - 2 15}{10}

Factor

- 10 - 215 : - 2 (5 + 15)

- 10 - 215

Rewrite as

= - 2 \cdot 5 - 215

Factor out common term

2

= - 2 (5 + 15)

= - \frac{2 ( 5 + 15 )}{10}

Cancel the common factor:

2

= - \frac{5 + 15}{5}

The solutions to the quadratic equation are:

u = \frac{- 5 + 15}{5}, u = - \frac{5 + 15}{5}

Substitute back

u = sin (x)

sin (x) = \frac{- 5 + 15}{5}, sin (x) = - \frac{5 + 15}{5}

sin (x) = \frac{- 5 + 15}{5}, sin (x) = - \frac{5 + 15}{5}

sin (x) = \frac{- 5 + 15}{5} : x = arcsin (\frac{- 5 + 15}{5}) + 2 πn, x = π + arcsin (- \frac{- 5 + 15}{5}) + 2 πn

sin (x) = \frac{- 5 + 15}{5}

Apply trig inverse properties

sin (x) = \frac{- 5 + 15}{5}

General solutions for

sin (x) = \frac{- 5 + 15}{5}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{- 5 + 15}{5}) + 2 πn, x = π + arcsin (- \frac{- 5 + 15}{5}) + 2 πn

x = arcsin (\frac{- 5 + 15}{5}) + 2 πn, x = π + arcsin (- \frac{- 5 + 15}{5}) + 2 πn

sin (x) = - \frac{5 + 15}{5} :

No Solution

sin (x) = - \frac{5 + 15}{5}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (\frac{- 5 + 15}{5}) + 2 πn, x = π + arcsin (- \frac{- 5 + 15}{5}) + 2 πn

Show solutions in decimal form

x = - 0.22735 \dots + 2 πn, x = π + 0.22735 \dots + 2 πn

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tan (x) = \frac{5}{5.1}

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Frequently Asked Questions (FAQ)

What is the general solution for 5sin^2(x)+10sin(x)+2=0 ?
The general solution for 5sin^2(x)+10sin(x)+2=0 is x=-0.22735…+2pin,x=pi+0.22735…+2pin