What is the general solution for 8sin(x)=cos(x)-6 ?

The general solution for 8sin(x)=cos(x)-6 is x=-2.17788…+2pin,x=2pi-0.71499…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

8sin(x)=cos(x)-6

Solution

8 sin (x) = cos (x) - 6

Solution

x = - 2.17788 \dots + 2 πn, x = 2 π - 0.71499 \dots + 2 πn

Degrees

x = - 124.78383 \dots^{\circ} + 36 0^{\circ} n, x = 319.03386 \dots^{\circ} + 36 0^{\circ} n

Solution steps

8 sin (x) = cos (x) - 6

Square both sides

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

Subtract

(cos (x) - 6)^{2}

from both sides

64 sin^{2} (x) - cos^{2} (x) + 12 cos (x) - 36 = 0

Rewrite using trig identities

- 36 - cos^{2} (x) + 12 cos (x) + 64 sin^{2} (x)

Use the Pythagorean identity:

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

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

= - 36 - cos^{2} (x) + 12 cos (x) + 64 (1 - cos^{2} (x))

Simplify

- 36 - cos^{2} (x) + 12 cos (x) + 64 (1 - cos^{2} (x)) : 12 cos (x) - 65 cos^{2} (x) + 28

- 36 - cos^{2} (x) + 12 cos (x) + 64 (1 - cos^{2} (x))

Expand

64 (1 - cos^{2} (x)) : 64 - 64 cos^{2} (x)

64 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 64, b = 1, c = cos^{2} (x)

= 64 \cdot 1 - 64 cos^{2} (x)

Multiply the numbers:

64 \cdot 1 = 64

= 64 - 64 cos^{2} (x)

= - 36 - cos^{2} (x) + 12 cos (x) + 64 - 64 cos^{2} (x)

Simplify

- 36 - cos^{2} (x) + 12 cos (x) + 64 - 64 cos^{2} (x) : 12 cos (x) - 65 cos^{2} (x) + 28

- 36 - cos^{2} (x) + 12 cos (x) + 64 - 64 cos^{2} (x)

Group like terms

= - cos^{2} (x) + 12 cos (x) - 64 cos^{2} (x) - 36 + 64

Add similar elements:

- cos^{2} (x) - 64 cos^{2} (x) = - 65 cos^{2} (x)

= - 65 cos^{2} (x) + 12 cos (x) - 36 + 64

Add/Subtract the numbers:

- 36 + 64 = 28

= 12 cos (x) - 65 cos^{2} (x) + 28

= 12 cos (x) - 65 cos^{2} (x) + 28

= 12 cos (x) - 65 cos^{2} (x) + 28

28 + 12 cos (x) - 65 cos^{2} (x) = 0

Solve by substitution

28 + 12 cos (x) - 65 cos^{2} (x) = 0

Let:

cos (x) = u

28 + 12 u - 65 u^{2} = 0

28 + 12 u - 65 u^{2} = 0 : u = - \frac{2 ( 4 29 - 3 )}{65}, u = \frac{2 ( 3 + 4 29 )}{65}

28 + 12 u - 65 u^{2} = 0

Write in the standard form

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

- 65 u^{2} + 12 u + 28 = 0

Solve with the quadratic formula

- 65 u^{2} + 12 u + 28 = 0

Quadratic Equation Formula:

For

a = - 65, b = 12, c = 28

u_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 65 ) \cdot 28}{2 ( - 65 )}

u_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 65 ) \cdot 28}{2 ( - 65 )}

1 2^{2} - 4 (- 65) \cdot 28 = 1629

1 2^{2} - 4 (- 65) \cdot 28

Apply rule

- (- a) = a

= 1 2^{2} + 4 \cdot 65 \cdot 28

Multiply the numbers:

4 \cdot 65 \cdot 28 = 7280

= 1 2^{2} + 7280

1 2^{2} = 144

= 144 + 7280

Add the numbers:

144 + 7280 = 7424

= 7424

Prime factorization of

7424 : 2^{8} \cdot 29

7424

7424

divides by

27424 = 3712 \cdot 2

= 2 \cdot 3712

3712

divides by

23712 = 1856 \cdot 2

= 2 \cdot 2 \cdot 1856

1856

divides by

21856 = 928 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 928

928

divides by

2928 = 464 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 464

464

divides by

2464 = 232 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 232

232

divides by

2232 = 116 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 116

116

divides by

2116 = 58 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 58

58

divides by

258 = 29 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 29

2, 29

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 29

= 2^{8} \cdot 29

= 2^{8} \cdot 29

Apply radical rule:

n ab = n a n b

= 29 2^{8}

Apply radical rule:

n a^{m} = a^{\frac{m}{n}}

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

= 2^{4} 29

Refine

= 1629

u_{1, 2} = \frac{- 12 \pm 16 29}{2 ( - 65 )}

Separate the solutions

u_{1} = \frac{- 12 + 16 29}{2 ( - 65 )}, u_{2} = \frac{- 12 - 16 29}{2 ( - 65 )}

u = \frac{- 12 + 16 29}{2 ( - 65 )} : - \frac{2 ( 4 29 - 3 )}{65}

\frac{- 12 + 16 29}{2 ( - 65 )}

Remove parentheses:

(- a) = - a

= \frac{- 12 + 16 29}{- 2 \cdot 65}

Multiply the numbers:

2 \cdot 65 = 130

= \frac{- 12 + 16 29}{- 130}

Apply the fraction rule:

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

= - \frac{- 12 + 16 29}{130}

Cancel

\frac{- 12 + 16 29}{130} : \frac{2 ( 4 29 - 3 )}{65}

\frac{- 12 + 16 29}{130}

Factor

- 12 + 1629 : 4 (- 3 + 429)

- 12 + 1629

Rewrite as

= - 4 \cdot 3 + 4 \cdot 429

Factor out common term

4

= 4 (- 3 + 429)

= \frac{4 ( - 3 + 4 29 )}{130}

Cancel the common factor:

2

= \frac{2 ( 4 29 - 3 )}{65}

= - \frac{2 ( 4 29 - 3 )}{65}

u = \frac{- 12 - 16 29}{2 ( - 65 )} : \frac{2 ( 3 + 4 29 )}{65}

\frac{- 12 - 16 29}{2 ( - 65 )}

Remove parentheses:

(- a) = - a

= \frac{- 12 - 16 29}{- 2 \cdot 65}

Multiply the numbers:

2 \cdot 65 = 130

= \frac{- 12 - 16 29}{- 130}

Apply the fraction rule:

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

- 12 - 1629 = - (12 + 1629)

= \frac{12 + 16 29}{130}

Factor

12 + 1629 : 4 (3 + 429)

12 + 1629

Rewrite as

= 4 \cdot 3 + 4 \cdot 429

Factor out common term

4

= 4 (3 + 429)

= \frac{4 ( 3 + 4 29 )}{130}

Cancel the common factor:

2

= \frac{2 ( 3 + 4 29 )}{65}

The solutions to the quadratic equation are:

u = - \frac{2 ( 4 29 - 3 )}{65}, u = \frac{2 ( 3 + 4 29 )}{65}

Substitute back

u = cos (x)

cos (x) = - \frac{2 ( 4 29 - 3 )}{65}, cos (x) = \frac{2 ( 3 + 4 29 )}{65}

cos (x) = - \frac{2 ( 4 29 - 3 )}{65}, cos (x) = \frac{2 ( 3 + 4 29 )}{65}

cos (x) = - \frac{2 ( 4 29 - 3 )}{65} : x = arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn, x = - arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn

cos (x) = - \frac{2 ( 4 29 - 3 )}{65}

Apply trig inverse properties

cos (x) = - \frac{2 ( 4 29 - 3 )}{65}

General solutions for

cos (x) = - \frac{2 ( 4 29 - 3 )}{65}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn, x = - arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn

x = arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn, x = - arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn

cos (x) = \frac{2 ( 3 + 4 29 )}{65} : x = arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

cos (x) = \frac{2 ( 3 + 4 29 )}{65}

Apply trig inverse properties

cos (x) = \frac{2 ( 3 + 4 29 )}{65}

General solutions for

cos (x) = \frac{2 ( 3 + 4 29 )}{65}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

x = arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

Combine all the solutions

x = arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn, x = - arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn, x = arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

8 sin (x) = cos (x) - 6

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

Check the solution

arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn :

False

arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn

Plug in

n = 1

arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1

For

8 sin (x) = cos (x) - 6

plug in

x = arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1

8 sin (arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1) = cos (arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1) - 6

Refine

6.57048 \dots = - 6.57048 \dots

\Rightarrow False

Check the solution

- arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn :

True

- arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn

Plug in

n = 1

- arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1

For

8 sin (x) = cos (x) - 6

plug in

x = - arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1

8 sin (- arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1) = cos (- arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 π 1) - 6

Refine

- 6.57048 \dots = - 6.57048 \dots

\Rightarrow True

Check the solution

arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn :

False

arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

Plug in

n = 1

arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1

For

8 sin (x) = cos (x) - 6

plug in

x = arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1

8 sin (arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1) = cos (arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1) - 6

Refine

5.24490 \dots = - 5.24490 \dots

\Rightarrow False

Check the solution

2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn :

True

2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1

For

8 sin (x) = cos (x) - 6

plug in

x = 2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1

8 sin (2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1) = cos (2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 π 1) - 6

Refine

- 5.24490 \dots = - 5.24490 \dots

\Rightarrow True

x = - arccos (- \frac{2 ( 4 29 - 3 )}{65}) + 2 πn, x = 2 π - arccos (\frac{2 ( 3 + 4 29 )}{65}) + 2 πn

Show solutions in decimal form

x = - 2.17788 \dots + 2 πn, x = 2 π - 0.71499 \dots + 2 πn

Graph

Popular Examples

27sin(t)cos(t)=-3sin(t)

27 sin (t) cos (t) = - 3 sin (t)

sin(5x)-sin(x)+3cos(3x)=0

sin (5 x) - sin (x) + 3 cos (3 x) = 0

2+2sin(θ)=cos^2(θ)

2 + 2 sin (θ) = cos^{2} (θ)

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

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

-19sin(θ)+19/2 =0

- 19 sin (θ) + \frac{19}{2} = 0

Frequently Asked Questions (FAQ)

What is the general solution for 8sin(x)=cos(x)-6 ?
The general solution for 8sin(x)=cos(x)-6 is x=-2.17788…+2pin,x=2pi-0.71499…+2pin