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

The general solution for 5sin(x)=cos(x)-4 is x=-2.04236…+2pin,x=2pi-0.70443…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = - 2.04236 \dots + 2 πn, x = 2 π - 0.70443 \dots + 2 πn

Degrees

x = - 117.01888 \dots^{\circ} + 36 0^{\circ} n, x = 319.63875 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Square both sides

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

Subtract

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

from both sides

25 sin^{2} (x) - cos^{2} (x) + 8 cos (x) - 16 = 0

Rewrite using trig identities

- 16 - cos^{2} (x) + 25 sin^{2} (x) + 8 cos (x)

Use the Pythagorean identity:

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

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

= - 16 - cos^{2} (x) + 25 (1 - cos^{2} (x)) + 8 cos (x)

Simplify

- 16 - cos^{2} (x) + 25 (1 - cos^{2} (x)) + 8 cos (x) : 8 cos (x) - 26 cos^{2} (x) + 9

- 16 - cos^{2} (x) + 25 (1 - cos^{2} (x)) + 8 cos (x)

Expand

25 (1 - cos^{2} (x)) : 25 - 25 cos^{2} (x)

25 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 25, b = 1, c = cos^{2} (x)

= 25 \cdot 1 - 25 cos^{2} (x)

Multiply the numbers:

25 \cdot 1 = 25

= 25 - 25 cos^{2} (x)

= - 16 - cos^{2} (x) + 25 - 25 cos^{2} (x) + 8 cos (x)

Simplify

- 16 - cos^{2} (x) + 25 - 25 cos^{2} (x) + 8 cos (x) : 8 cos (x) - 26 cos^{2} (x) + 9

- 16 - cos^{2} (x) + 25 - 25 cos^{2} (x) + 8 cos (x)

Group like terms

= - cos^{2} (x) - 25 cos^{2} (x) + 8 cos (x) - 16 + 25

Add similar elements:

- cos^{2} (x) - 25 cos^{2} (x) = - 26 cos^{2} (x)

= - 26 cos^{2} (x) + 8 cos (x) - 16 + 25

Add/Subtract the numbers:

- 16 + 25 = 9

= 8 cos (x) - 26 cos^{2} (x) + 9

= 8 cos (x) - 26 cos^{2} (x) + 9

= 8 cos (x) - 26 cos^{2} (x) + 9

9 - 26 cos^{2} (x) + 8 cos (x) = 0

Solve by substitution

9 - 26 cos^{2} (x) + 8 cos (x) = 0

Let:

cos (x) = u

9 - 26 u^{2} + 8 u = 0

9 - 26 u^{2} + 8 u = 0 : u = - \frac{- 4 + 5 10}{26}, u = \frac{4 + 5 10}{26}

9 - 26 u^{2} + 8 u = 0

Write in the standard form

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

- 26 u^{2} + 8 u + 9 = 0

Solve with the quadratic formula

- 26 u^{2} + 8 u + 9 = 0

Quadratic Equation Formula:

For

a = - 26, b = 8, c = 9

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 26 ) \cdot 9}{2 ( - 26 )}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 26 ) \cdot 9}{2 ( - 26 )}

8^{2} - 4 (- 26) \cdot 9 = 1010

8^{2} - 4 (- 26) \cdot 9

Apply rule

- (- a) = a

= 8^{2} + 4 \cdot 26 \cdot 9

Multiply the numbers:

4 \cdot 26 \cdot 9 = 936

= 8^{2} + 936

8^{2} = 64

= 64 + 936

Add the numbers:

64 + 936 = 1000

= 1000

Prime factorization of

1000 : 2^{3} \cdot 5^{3}

1000

1000

divides by

21000 = 500 \cdot 2

= 2 \cdot 500

500

divides by

2500 = 250 \cdot 2

= 2 \cdot 2 \cdot 250

250

divides by

2250 = 125 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 125

125

divides by

5125 = 25 \cdot 5

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 25

25

divides by

525 = 5 \cdot 5

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5

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

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

Apply exponent rule:

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

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

Apply radical rule:

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

Apply radical rule:

2^{2} = 2

= 2 5^{2} 2 \cdot 5

Apply radical rule:

5^{2} = 5

= 2 \cdot 5 2 \cdot 5

Refine

= 1010

u_{1, 2} = \frac{- 8 \pm 10 10}{2 ( - 26 )}

Separate the solutions

u_{1} = \frac{- 8 + 10 10}{2 ( - 26 )}, u_{2} = \frac{- 8 - 10 10}{2 ( - 26 )}

u = \frac{- 8 + 10 10}{2 ( - 26 )} : - \frac{- 4 + 5 10}{26}

\frac{- 8 + 10 10}{2 ( - 26 )}

Remove parentheses:

(- a) = - a

= \frac{- 8 + 10 10}{- 2 \cdot 26}

Multiply the numbers:

2 \cdot 26 = 52

= \frac{- 8 + 10 10}{- 52}

Apply the fraction rule:

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

= - \frac{- 8 + 10 10}{52}

Cancel

\frac{- 8 + 10 10}{52} : \frac{5 10 - 4}{26}

\frac{- 8 + 10 10}{52}

Factor

- 8 + 1010 : 2 (- 4 + 510)

- 8 + 1010

Rewrite as

= - 2 \cdot 4 + 2 \cdot 510

Factor out common term

2

= 2 (- 4 + 510)

= \frac{2 ( - 4 + 5 10 )}{52}

Cancel the common factor:

2

= \frac{- 4 + 5 10}{26}

= - \frac{5 10 - 4}{26}

= - \frac{- 4 + 5 10}{26}

u = \frac{- 8 - 10 10}{2 ( - 26 )} : \frac{4 + 5 10}{26}

\frac{- 8 - 10 10}{2 ( - 26 )}

Remove parentheses:

(- a) = - a

= \frac{- 8 - 10 10}{- 2 \cdot 26}

Multiply the numbers:

2 \cdot 26 = 52

= \frac{- 8 - 10 10}{- 52}

Apply the fraction rule:

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

- 8 - 1010 = - (8 + 1010)

= \frac{8 + 10 10}{52}

Factor

8 + 1010 : 2 (4 + 510)

8 + 1010

Rewrite as

= 2 \cdot 4 + 2 \cdot 510

Factor out common term

2

= 2 (4 + 510)

= \frac{2 ( 4 + 5 10 )}{52}

Cancel the common factor:

2

= \frac{4 + 5 10}{26}

The solutions to the quadratic equation are:

u = - \frac{- 4 + 5 10}{26}, u = \frac{4 + 5 10}{26}

Substitute back

u = cos (x)

cos (x) = - \frac{- 4 + 5 10}{26}, cos (x) = \frac{4 + 5 10}{26}

cos (x) = - \frac{- 4 + 5 10}{26}, cos (x) = \frac{4 + 5 10}{26}

cos (x) = - \frac{- 4 + 5 10}{26} : x = arccos (- \frac{- 4 + 5 10}{26}) + 2 πn, x = - arccos (- \frac{- 4 + 5 10}{26}) + 2 πn

cos (x) = - \frac{- 4 + 5 10}{26}

Apply trig inverse properties

cos (x) = - \frac{- 4 + 5 10}{26}

General solutions for

cos (x) = - \frac{- 4 + 5 10}{26}

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

x = arccos (- \frac{- 4 + 5 10}{26}) + 2 πn, x = - arccos (- \frac{- 4 + 5 10}{26}) + 2 πn

x = arccos (- \frac{- 4 + 5 10}{26}) + 2 πn, x = - arccos (- \frac{- 4 + 5 10}{26}) + 2 πn

cos (x) = \frac{4 + 5 10}{26} : x = arccos (\frac{4 + 5 10}{26}) + 2 πn, x = 2 π - arccos (\frac{4 + 5 10}{26}) + 2 πn

cos (x) = \frac{4 + 5 10}{26}

Apply trig inverse properties

cos (x) = \frac{4 + 5 10}{26}

General solutions for

cos (x) = \frac{4 + 5 10}{26}

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

x = arccos (\frac{4 + 5 10}{26}) + 2 πn, x = 2 π - arccos (\frac{4 + 5 10}{26}) + 2 πn

x = arccos (\frac{4 + 5 10}{26}) + 2 πn, x = 2 π - arccos (\frac{4 + 5 10}{26}) + 2 πn

Combine all the solutions

x = arccos (- \frac{- 4 + 5 10}{26}) + 2 πn, x = - arccos (- \frac{- 4 + 5 10}{26}) + 2 πn, x = arccos (\frac{4 + 5 10}{26}) + 2 πn, x = 2 π - arccos (\frac{4 + 5 10}{26}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

arccos (- \frac{- 4 + 5 10}{26}) + 2 πn :

False

arccos (- \frac{- 4 + 5 10}{26}) + 2 πn

Plug in

n = 1

arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1

For

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

plug in

x = arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1

5 sin (arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1) = cos (arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1) - 4

Refine

4.45428 \dots = - 4.45428 \dots

\Rightarrow False

Check the solution

- arccos (- \frac{- 4 + 5 10}{26}) + 2 πn :

True

- arccos (- \frac{- 4 + 5 10}{26}) + 2 πn

Plug in

n = 1

- arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1

For

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

plug in

x = - arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1

5 sin (- arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1) = cos (- arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1) - 4

Refine

- 4.45428 \dots = - 4.45428 \dots

\Rightarrow True

Check the solution

arccos (\frac{4 + 5 10}{26}) + 2 πn :

False

arccos (\frac{4 + 5 10}{26}) + 2 πn

Plug in

n = 1

arccos (\frac{4 + 5 10}{26}) + 2 π 1

For

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

plug in

x = arccos (\frac{4 + 5 10}{26}) + 2 π 1

5 sin (arccos (\frac{4 + 5 10}{26}) + 2 π 1) = cos (arccos (\frac{4 + 5 10}{26}) + 2 π 1) - 4

Refine

3.23802 \dots = - 3.23802 \dots

\Rightarrow False

Check the solution

2 π - arccos (\frac{4 + 5 10}{26}) + 2 πn :

True

2 π - arccos (\frac{4 + 5 10}{26}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{4 + 5 10}{26}) + 2 π 1

For

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

plug in

x = 2 π - arccos (\frac{4 + 5 10}{26}) + 2 π 1

5 sin (2 π - arccos (\frac{4 + 5 10}{26}) + 2 π 1) = cos (2 π - arccos (\frac{4 + 5 10}{26}) + 2 π 1) - 4

Refine

- 3.23802 \dots = - 3.23802 \dots

\Rightarrow True

x = - arccos (- \frac{- 4 + 5 10}{26}) + 2 πn, x = 2 π - arccos (\frac{4 + 5 10}{26}) + 2 πn

Show solutions in decimal form

x = - 2.04236 \dots + 2 πn, x = 2 π - 0.70443 \dots + 2 πn

Popular Examples

tan(x)-3=0

tan (x) - 3 = 0

solvefor x,2sin^2(x)-sin(x)-1=0 solve for

x, 2 sin^{2} (x) - sin (x) - 1 = 0

-21tan(x)+3sqrt(3)=-3(sqrt(3)+tan(x))

- 21 tan (x) + 33 = - 3 (3 + tan (x))

tan(y)=-1

tan (y) = - 1

2pi+6arccos(2-x)=0

2 π + 6 arccos (2 - x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 5sin(x)=cos(x)-4 ?
The general solution for 5sin(x)=cos(x)-4 is x=-2.04236…+2pin,x=2pi-0.70443…+2pin