What is the general solution for sin(x)+4cos(x)+5=0 ?

The general solution for sin(x)+4cos(x)+5=0 is No Solution for x\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

sin(x)+4cos(x)+5=0

Solution

sin (x) + 4 cos (x) + 5 = 0

Solution

No Solution for x \in R

Solution steps

sin (x) + 4 cos (x) + 5 = 0

Subtract

4 cos (x)

from both sides

sin (x) + 5 = - 4 cos (x)

Square both sides

(sin (x) + 5)^{2} = (- 4 cos (x))^{2}

Subtract

(- 4 cos (x))^{2}

from both sides

(sin (x) + 5)^{2} - 16 cos^{2} (x) = 0

Rewrite using trig identities

(5 + sin (x))^{2} - 16 cos^{2} (x)

Use the Pythagorean identity:

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

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

= (5 + sin (x))^{2} - 16 (1 - sin^{2} (x))

Simplify

(5 + sin (x))^{2} - 16 (1 - sin^{2} (x)) : 17 sin^{2} (x) + 10 sin (x) + 9

(5 + sin (x))^{2} - 16 (1 - sin^{2} (x))

(5 + sin (x))^{2} : 25 + 10 sin (x) + sin^{2} (x)

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 5, b = sin (x)

= 5^{2} + 2 \cdot 5 sin (x) + sin^{2} (x)

Simplify

5^{2} + 2 \cdot 5 sin (x) + sin^{2} (x) : 25 + 10 sin (x) + sin^{2} (x)

5^{2} + 2 \cdot 5 sin (x) + sin^{2} (x)

5^{2} = 25

= 25 + 2 \cdot 5 sin (x) + sin^{2} (x)

Multiply the numbers:

2 \cdot 5 = 10

= 25 + 10 sin (x) + sin^{2} (x)

= 25 + 10 sin (x) + sin^{2} (x)

= 25 + 10 sin (x) + sin^{2} (x) - 16 (1 - sin^{2} (x))

Expand

- 16 (1 - sin^{2} (x)) : - 16 + 16 sin^{2} (x)

- 16 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 16, b = 1, c = sin^{2} (x)

= - 16 \cdot 1 - (- 16) sin^{2} (x)

Apply minus-plus rules

- (- a) = a

= - 16 \cdot 1 + 16 sin^{2} (x)

Multiply the numbers:

16 \cdot 1 = 16

= - 16 + 16 sin^{2} (x)

= 25 + 10 sin (x) + sin^{2} (x) - 16 + 16 sin^{2} (x)

Simplify

25 + 10 sin (x) + sin^{2} (x) - 16 + 16 sin^{2} (x) : 17 sin^{2} (x) + 10 sin (x) + 9

25 + 10 sin (x) + sin^{2} (x) - 16 + 16 sin^{2} (x)

Group like terms

= 10 sin (x) + sin^{2} (x) + 16 sin^{2} (x) + 25 - 16

Add similar elements:

sin^{2} (x) + 16 sin^{2} (x) = 17 sin^{2} (x)

= 10 sin (x) + 17 sin^{2} (x) + 25 - 16

Add/Subtract the numbers:

25 - 16 = 9

= 17 sin^{2} (x) + 10 sin (x) + 9

= 17 sin^{2} (x) + 10 sin (x) + 9

= 17 sin^{2} (x) + 10 sin (x) + 9

9 + 10 sin (x) + 17 sin^{2} (x) = 0

Solve by substitution

9 + 10 sin (x) + 17 sin^{2} (x) = 0

Let:

sin (x) = u

9 + 10 u + 17 u^{2} = 0

9 + 10 u + 17 u^{2} = 0 : u = - \frac{5}{17} + i \frac{8 2}{17}, u = - \frac{5}{17} - i \frac{8 2}{17}

9 + 10 u + 17 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

17 u^{2} + 10 u + 9 = 0

Solve with the quadratic formula

17 u^{2} + 10 u + 9 = 0

Quadratic Equation Formula:

For

a = 17, b = 10, c = 9

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 17 \cdot 9}{2 \cdot 17}

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 17 \cdot 9}{2 \cdot 17}

Simplify

1 0^{2} - 4 \cdot 17 \cdot 9 : 162 i

1 0^{2} - 4 \cdot 17 \cdot 9

Multiply the numbers:

4 \cdot 17 \cdot 9 = 612

= 1 0^{2} - 612

Apply imaginary number rule:

- a = i a

= i 612 - 1 0^{2}

- 1 0^{2} + 612 = 162

- 1 0^{2} + 612

1 0^{2} = 100

= - 100 + 612

Add/Subtract the numbers:

- 100 + 612 = 512

= 512

Prime factorization of

512 : 2^{9}

512

512

divides by

2512 = 256 \cdot 2

= 2 \cdot 256

256

divides by

2256 = 128 \cdot 2

= 2 \cdot 2 \cdot 128

128

divides by

2128 = 64 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 64

64

divides by

264 = 32 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 32

32

divides by

232 = 16 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{9}

= 2^{9}

Apply exponent rule:

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

= 2^{8} \cdot 2

Apply radical rule:

= 2 2^{8}

Apply radical rule:

2^{8} = 2^{\frac{8}{2}} = 2^{4}

= 2^{4} 2

Refine

= 162

= 162 i

u_{1, 2} = \frac{- 10 \pm 16 2 i}{2 \cdot 17}

Separate the solutions

u_{1} = \frac{- 10 + 16 2 i}{2 \cdot 17}, u_{2} = \frac{- 10 - 16 2 i}{2 \cdot 17}

u = \frac{- 10 + 16 2 i}{2 \cdot 17} : - \frac{5}{17} + i \frac{8 2}{17}

\frac{- 10 + 16 2 i}{2 \cdot 17}

Multiply the numbers:

2 \cdot 17 = 34

= \frac{- 10 + 16 2 i}{34}

Factor

- 10 + 162 i : 2 (- 5 + 82 i)

- 10 + 162 i

Rewrite as

= - 2 \cdot 5 + 2 \cdot 82 i

Factor out common term

2

= 2 (- 5 + 82 i)

= \frac{2 ( - 5 + 8 2 i )}{34}

Cancel the common factor:

2

= \frac{- 5 + 8 2 i}{17}

Rewrite

\frac{- 5 + 8 2 i}{17}

in standard complex form:

- \frac{5}{17} + \frac{8 2}{17} i

\frac{- 5 + 8 2 i}{17}

Apply the fraction rule:

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

\frac{- 5 + 8 2 i}{17} = - \frac{5}{17} + \frac{8 2 i}{17}

= - \frac{5}{17} + \frac{8 2 i}{17}

= - \frac{5}{17} + \frac{8 2}{17} i

u = \frac{- 10 - 16 2 i}{2 \cdot 17} : - \frac{5}{17} - i \frac{8 2}{17}

\frac{- 10 - 16 2 i}{2 \cdot 17}

Multiply the numbers:

2 \cdot 17 = 34

= \frac{- 10 - 16 2 i}{34}

Factor

- 10 - 162 i : - 2 (5 + 82 i)

- 10 - 162 i

Rewrite as

= - 2 \cdot 5 - 2 \cdot 82 i

Factor out common term

2

= - 2 (5 + 82 i)

= - \frac{2 ( 5 + 8 2 i )}{34}

Cancel the common factor:

2

= - \frac{5 + 8 2 i}{17}

Rewrite

- \frac{5 + 8 2 i}{17}

in standard complex form:

- \frac{5}{17} - \frac{8 2}{17} i

- \frac{5 + 8 2 i}{17}

Apply the fraction rule:

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

\frac{5 + 8 2 i}{17} = - (\frac{5}{17}) - (\frac{8 2 i}{17})

= - (\frac{5}{17}) - (\frac{8 2 i}{17})

Remove parentheses:

(a) = a

= - \frac{5}{17} - \frac{8 2 i}{17}

= - \frac{5}{17} - \frac{8 2}{17} i

The solutions to the quadratic equation are:

u = - \frac{5}{17} + i \frac{8 2}{17}, u = - \frac{5}{17} - i \frac{8 2}{17}

Substitute back

u = sin (x)

sin (x) = - \frac{5}{17} + i \frac{8 2}{17}, sin (x) = - \frac{5}{17} - i \frac{8 2}{17}

sin (x) = - \frac{5}{17} + i \frac{8 2}{17}, sin (x) = - \frac{5}{17} - i \frac{8 2}{17}

sin (x) = - \frac{5}{17} + i \frac{8 2}{17} :

No Solution

sin (x) = - \frac{5}{17} + i \frac{8 2}{17}

No Solution

sin (x) = - \frac{5}{17} - i \frac{8 2}{17} :

No Solution

sin (x) = - \frac{5}{17} - i \frac{8 2}{17}

No Solution

Combine all the solutions

No Solution

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

sin (x) + 4 cos (x) + 5 = 0

Remove the ones that don't agree with the equation.

No Solution for x \in R

Graph

Popular Examples

sec(x)=-5

sin(2t-pi/5)=-1/3

2(sin(x))^2-cos(x)-1=0

arctan(3x)+arctan(x)= pi/4

4sin(θ)=sqrt(3)sec(θ),0<= θ<180

Frequently Asked Questions (FAQ)

What is the general solution for sin(x)+4cos(x)+5=0 ?
The general solution for sin(x)+4cos(x)+5=0 is No Solution for x\in\mathbb{R}