What is the general solution for 8sin(x)=cos(x)-3 ?

The general solution for 8sin(x)=cos(x)-3 is x=-2.63596…+2pin,x=2pi-0.25691…+2pin

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Popular Trigonometry >

8sin(x)=cos(x)-3

Solution

8 sin (x) = cos (x) - 3

Solution

x = - 2.63596 \dots + 2 πn, x = 2 π - 0.25691 \dots + 2 πn

Degrees

x = - 151.02953 \dots^{\circ} + 36 0^{\circ} n, x = 345.27956 \dots^{\circ} + 36 0^{\circ} n

Solution steps

8 sin (x) = cos (x) - 3

Square both sides

(8 sin (x))^{2} = (cos (x) - 3)^{2}

Subtract

(cos (x) - 3)^{2}

from both sides

64 sin^{2} (x) - cos^{2} (x) + 6 cos (x) - 9 = 0

Rewrite using trig identities

- 9 - cos^{2} (x) + 64 sin^{2} (x) + 6 cos (x)

Use the Pythagorean identity:

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

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

= - 9 - cos^{2} (x) + 64 (1 - cos^{2} (x)) + 6 cos (x)

Simplify

- 9 - cos^{2} (x) + 64 (1 - cos^{2} (x)) + 6 cos (x) : 6 cos (x) - 65 cos^{2} (x) + 55

- 9 - cos^{2} (x) + 64 (1 - cos^{2} (x)) + 6 cos (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)

= - 9 - cos^{2} (x) + 64 - 64 cos^{2} (x) + 6 cos (x)

Simplify

- 9 - cos^{2} (x) + 64 - 64 cos^{2} (x) + 6 cos (x) : 6 cos (x) - 65 cos^{2} (x) + 55

- 9 - cos^{2} (x) + 64 - 64 cos^{2} (x) + 6 cos (x)

Group like terms

= - cos^{2} (x) - 64 cos^{2} (x) + 6 cos (x) - 9 + 64

Add similar elements:

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

= - 65 cos^{2} (x) + 6 cos (x) - 9 + 64

Add/Subtract the numbers:

- 9 + 64 = 55

= 6 cos (x) - 65 cos^{2} (x) + 55

= 6 cos (x) - 65 cos^{2} (x) + 55

= 6 cos (x) - 65 cos^{2} (x) + 55

55 - 65 cos^{2} (x) + 6 cos (x) = 0

Solve by substitution

55 - 65 cos^{2} (x) + 6 cos (x) = 0

Let:

cos (x) = u

55 - 65 u^{2} + 6 u = 0

55 - 65 u^{2} + 6 u = 0 : u = - \frac{- 3 + 16 14}{65}, u = \frac{3 + 16 14}{65}

55 - 65 u^{2} + 6 u = 0

Write in the standard form

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

- 65 u^{2} + 6 u + 55 = 0

Solve with the quadratic formula

- 65 u^{2} + 6 u + 55 = 0

Quadratic Equation Formula:

For

a = - 65, b = 6, c = 55

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 65 ) \cdot 55}{2 ( - 65 )}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 65 ) \cdot 55}{2 ( - 65 )}

6^{2} - 4 (- 65) \cdot 55 = 3214

6^{2} - 4 (- 65) \cdot 55

Apply rule

- (- a) = a

= 6^{2} + 4 \cdot 65 \cdot 55

Multiply the numbers:

4 \cdot 65 \cdot 55 = 14300

= 6^{2} + 14300

6^{2} = 36

= 36 + 14300

Add the numbers:

36 + 14300 = 14336

= 14336

Prime factorization of

14336 : 2^{11} \cdot 7

14336

14336

divides by

214336 = 7168 \cdot 2

= 2 \cdot 7168

7168

divides by

27168 = 3584 \cdot 2

= 2 \cdot 2 \cdot 3584

3584

divides by

23584 = 1792 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 1792

1792

divides by

21792 = 896 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 896

896

divides by

2896 = 448 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 448

448

divides by

2448 = 224 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 224

224

divides by

2224 = 112 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 112

112

divides by

2112 = 56 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 56

56

divides by

256 = 28 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 28

28

divides by

228 = 14 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 14

14

divides by

214 = 7 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 7

2, 7

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 2 \cdot 2 \cdot 2 \cdot 7

= 2^{11} \cdot 7

= 2^{11} \cdot 7

Apply exponent rule:

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

= 2^{10} \cdot 2 \cdot 7

Apply radical rule:

= 2^{10} 2 \cdot 7

Apply radical rule:

2^{10} = 2^{\frac{10}{2}} = 2^{5}

= 2^{5} 2 \cdot 7

Refine

= 3214

u_{1, 2} = \frac{- 6 \pm 32 14}{2 ( - 65 )}

Separate the solutions

u_{1} = \frac{- 6 + 32 14}{2 ( - 65 )}, u_{2} = \frac{- 6 - 32 14}{2 ( - 65 )}

u = \frac{- 6 + 32 14}{2 ( - 65 )} : - \frac{- 3 + 16 14}{65}

\frac{- 6 + 32 14}{2 ( - 65 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 + 32 14}{- 2 \cdot 65}

Multiply the numbers:

2 \cdot 65 = 130

= \frac{- 6 + 32 14}{- 130}

Apply the fraction rule:

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

= - \frac{- 6 + 32 14}{130}

Cancel

\frac{- 6 + 32 14}{130} : \frac{16 14 - 3}{65}

\frac{- 6 + 32 14}{130}

Factor

- 6 + 3214 : 2 (- 3 + 1614)

- 6 + 3214

Rewrite as

= - 2 \cdot 3 + 2 \cdot 1614

Factor out common term

2

= 2 (- 3 + 1614)

= \frac{2 ( - 3 + 16 14 )}{130}

Cancel the common factor:

2

= \frac{- 3 + 16 14}{65}

= - \frac{16 14 - 3}{65}

= - \frac{- 3 + 16 14}{65}

u = \frac{- 6 - 32 14}{2 ( - 65 )} : \frac{3 + 16 14}{65}

\frac{- 6 - 32 14}{2 ( - 65 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 - 32 14}{- 2 \cdot 65}

Multiply the numbers:

2 \cdot 65 = 130

= \frac{- 6 - 32 14}{- 130}

Apply the fraction rule:

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

- 6 - 3214 = - (6 + 3214)

= \frac{6 + 32 14}{130}

Factor

6 + 3214 : 2 (3 + 1614)

6 + 3214

Rewrite as

= 2 \cdot 3 + 2 \cdot 1614

Factor out common term

2

= 2 (3 + 1614)

= \frac{2 ( 3 + 16 14 )}{130}

Cancel the common factor:

2

= \frac{3 + 16 14}{65}

The solutions to the quadratic equation are:

u = - \frac{- 3 + 16 14}{65}, u = \frac{3 + 16 14}{65}

Substitute back

u = cos (x)

cos (x) = - \frac{- 3 + 16 14}{65}, cos (x) = \frac{3 + 16 14}{65}

cos (x) = - \frac{- 3 + 16 14}{65}, cos (x) = \frac{3 + 16 14}{65}

cos (x) = - \frac{- 3 + 16 14}{65} : x = arccos (- \frac{- 3 + 16 14}{65}) + 2 πn, x = - arccos (- \frac{- 3 + 16 14}{65}) + 2 πn

cos (x) = - \frac{- 3 + 16 14}{65}

Apply trig inverse properties

cos (x) = - \frac{- 3 + 16 14}{65}

General solutions for

cos (x) = - \frac{- 3 + 16 14}{65}

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

x = arccos (- \frac{- 3 + 16 14}{65}) + 2 πn, x = - arccos (- \frac{- 3 + 16 14}{65}) + 2 πn

x = arccos (- \frac{- 3 + 16 14}{65}) + 2 πn, x = - arccos (- \frac{- 3 + 16 14}{65}) + 2 πn

cos (x) = \frac{3 + 16 14}{65} : x = arccos (\frac{3 + 16 14}{65}) + 2 πn, x = 2 π - arccos (\frac{3 + 16 14}{65}) + 2 πn

cos (x) = \frac{3 + 16 14}{65}

Apply trig inverse properties

cos (x) = \frac{3 + 16 14}{65}

General solutions for

cos (x) = \frac{3 + 16 14}{65}

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

x = arccos (\frac{3 + 16 14}{65}) + 2 πn, x = 2 π - arccos (\frac{3 + 16 14}{65}) + 2 πn

x = arccos (\frac{3 + 16 14}{65}) + 2 πn, x = 2 π - arccos (\frac{3 + 16 14}{65}) + 2 πn

Combine all the solutions

x = arccos (- \frac{- 3 + 16 14}{65}) + 2 πn, x = - arccos (- \frac{- 3 + 16 14}{65}) + 2 πn, x = arccos (\frac{3 + 16 14}{65}) + 2 πn, x = 2 π - arccos (\frac{3 + 16 14}{65}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

8 sin (x) = cos (x) - 3

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

Check the solution

arccos (- \frac{- 3 + 16 14}{65}) + 2 πn :

False

arccos (- \frac{- 3 + 16 14}{65}) + 2 πn

Plug in

n = 1

arccos (- \frac{- 3 + 16 14}{65}) + 2 π 1

For

8 sin (x) = cos (x) - 3

plug in

x = arccos (- \frac{- 3 + 16 14}{65}) + 2 π 1

8 sin (arccos (- \frac{- 3 + 16 14}{65}) + 2 π 1) = cos (arccos (- \frac{- 3 + 16 14}{65}) + 2 π 1) - 3

Refine

3.87486 \dots = - 3.87486 \dots

\Rightarrow False

Check the solution

- arccos (- \frac{- 3 + 16 14}{65}) + 2 πn :

True

- arccos (- \frac{- 3 + 16 14}{65}) + 2 πn

Plug in

n = 1

- arccos (- \frac{- 3 + 16 14}{65}) + 2 π 1

For

8 sin (x) = cos (x) - 3

plug in

x = - arccos (- \frac{- 3 + 16 14}{65}) + 2 π 1

8 sin (- arccos (- \frac{- 3 + 16 14}{65}) + 2 π 1) = cos (- arccos (- \frac{- 3 + 16 14}{65}) + 2 π 1) - 3

Refine

- 3.87486 \dots = - 3.87486 \dots

\Rightarrow True

Check the solution

arccos (\frac{3 + 16 14}{65}) + 2 πn :

False

arccos (\frac{3 + 16 14}{65}) + 2 πn

Plug in

n = 1

arccos (\frac{3 + 16 14}{65}) + 2 π 1

For

8 sin (x) = cos (x) - 3

plug in

x = arccos (\frac{3 + 16 14}{65}) + 2 π 1

8 sin (arccos (\frac{3 + 16 14}{65}) + 2 π 1) = cos (arccos (\frac{3 + 16 14}{65}) + 2 π 1) - 3

Refine

2.03282 \dots = - 2.03282 \dots

\Rightarrow False

Check the solution

2 π - arccos (\frac{3 + 16 14}{65}) + 2 πn :

True

2 π - arccos (\frac{3 + 16 14}{65}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{3 + 16 14}{65}) + 2 π 1

For

8 sin (x) = cos (x) - 3

plug in

x = 2 π - arccos (\frac{3 + 16 14}{65}) + 2 π 1

8 sin (2 π - arccos (\frac{3 + 16 14}{65}) + 2 π 1) = cos (2 π - arccos (\frac{3 + 16 14}{65}) + 2 π 1) - 3

Refine

- 2.03282 \dots = - 2.03282 \dots

\Rightarrow True

x = - arccos (- \frac{- 3 + 16 14}{65}) + 2 πn, x = 2 π - arccos (\frac{3 + 16 14}{65}) + 2 πn

Show solutions in decimal form

x = - 2.63596 \dots + 2 πn, x = 2 π - 0.25691 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 8sin(x)=cos(x)-3 ?
The general solution for 8sin(x)=cos(x)-3 is x=-2.63596…+2pin,x=2pi-0.25691…+2pin