What is the general solution for 5*cos(x)+4-2*sin(x)=0 ?

The general solution for 5*cos(x)+4-2*sin(x)=0 is x=2.02750…+2pin,x=-2.78851…+2pin

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Popular Trigonometry >

5cos(x)+4-2sin(x)=0

Solution

5 \cdot cos (x) + 4 - 2 \cdot sin (x) = 0

Solution

x = 2.02750 \dots + 2 πn, x = - 2.78851 \dots + 2 πn

Degrees

x = 116.16747 \dots^{\circ} + 36 0^{\circ} n, x = - 159.77029 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 cos (x) + 4 - 2 sin (x) = 0

Add

2 sin (x)

to both sides

5 cos (x) + 4 = 2 sin (x)

Square both sides

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

Subtract

(2 sin (x))^{2}

from both sides

(5 cos (x) + 4)^{2} - 4 sin^{2} (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= (4 + 5 cos (x))^{2} - 4 (1 - cos^{2} (x))

Simplify

(4 + 5 cos (x))^{2} - 4 (1 - cos^{2} (x)) : 29 cos^{2} (x) + 40 cos (x) + 12

(4 + 5 cos (x))^{2} - 4 (1 - cos^{2} (x))

(4 + 5 cos (x))^{2} : 16 + 40 cos (x) + 25 cos^{2} (x)

Apply Perfect Square Formula:

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

a = 4, b = 5 cos (x)

= 4^{2} + 2 \cdot 4 \cdot 5 cos (x) + (5 cos (x))^{2}

Simplify

4^{2} + 2 \cdot 4 \cdot 5 cos (x) + (5 cos (x))^{2} : 16 + 40 cos (x) + 25 cos^{2} (x)

4^{2} + 2 \cdot 4 \cdot 5 cos (x) + (5 cos (x))^{2}

4^{2} = 16

4^{2}

4^{2} = 16

= 16

2 \cdot 4 \cdot 5 cos (x) = 40 cos (x)

2 \cdot 4 \cdot 5 cos (x)

Multiply the numbers:

2 \cdot 4 \cdot 5 = 40

= 40 cos (x)

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

(5 cos (x))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 5^{2} cos^{2} (x)

5^{2} = 25

= 25 cos^{2} (x)

= 16 + 40 cos (x) + 25 cos^{2} (x)

= 16 + 40 cos (x) + 25 cos^{2} (x)

= 16 + 40 cos (x) + 25 cos^{2} (x) - 4 (1 - cos^{2} (x))

Expand

- 4 (1 - cos^{2} (x)) : - 4 + 4 cos^{2} (x)

- 4 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = cos^{2} (x)

= - 4 \cdot 1 - (- 4) cos^{2} (x)

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 cos^{2} (x)

Multiply the numbers:

4 \cdot 1 = 4

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

= 16 + 40 cos (x) + 25 cos^{2} (x) - 4 + 4 cos^{2} (x)

Simplify

16 + 40 cos (x) + 25 cos^{2} (x) - 4 + 4 cos^{2} (x) : 29 cos^{2} (x) + 40 cos (x) + 12

16 + 40 cos (x) + 25 cos^{2} (x) - 4 + 4 cos^{2} (x)

Group like terms

= 40 cos (x) + 25 cos^{2} (x) + 4 cos^{2} (x) + 16 - 4

Add similar elements:

25 cos^{2} (x) + 4 cos^{2} (x) = 29 cos^{2} (x)

= 40 cos (x) + 29 cos^{2} (x) + 16 - 4

Add/Subtract the numbers:

16 - 4 = 12

= 29 cos^{2} (x) + 40 cos (x) + 12

= 29 cos^{2} (x) + 40 cos (x) + 12

= 29 cos^{2} (x) + 40 cos (x) + 12

12 + 29 cos^{2} (x) + 40 cos (x) = 0

Solve by substitution

12 + 29 cos^{2} (x) + 40 cos (x) = 0

Let:

cos (x) = u

12 + 29 u^{2} + 40 u = 0

12 + 29 u^{2} + 40 u = 0 : u = \frac{2 ( 13 - 10 )}{29}, u = - \frac{2 ( 10 + 13 )}{29}

12 + 29 u^{2} + 40 u = 0

Write in the standard form

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

29 u^{2} + 40 u + 12 = 0

Solve with the quadratic formula

29 u^{2} + 40 u + 12 = 0

Quadratic Equation Formula:

For

a = 29, b = 40, c = 12

u_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 29 \cdot 12}{2 \cdot 29}

u_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 29 \cdot 12}{2 \cdot 29}

4 0^{2} - 4 \cdot 29 \cdot 12 = 413

4 0^{2} - 4 \cdot 29 \cdot 12

Multiply the numbers:

4 \cdot 29 \cdot 12 = 1392

= 4 0^{2} - 1392

4 0^{2} = 1600

= 1600 - 1392

Subtract the numbers:

1600 - 1392 = 208

= 208

Prime factorization of

208 : 2^{4} \cdot 13

208

208

divides by

2208 = 104 \cdot 2

= 2 \cdot 104

104

divides by

2104 = 52 \cdot 2

= 2 \cdot 2 \cdot 52

52

divides by

252 = 26 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 26

26

divides by

226 = 13 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 13

2, 13

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 13

= 2^{4} \cdot 13

= 2^{4} \cdot 13

Apply radical rule:

= 13 2^{4}

Apply radical rule:

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

= 2^{2} 13

Refine

= 413

u_{1, 2} = \frac{- 40 \pm 4 13}{2 \cdot 29}

Separate the solutions

u_{1} = \frac{- 40 + 4 13}{2 \cdot 29}, u_{2} = \frac{- 40 - 4 13}{2 \cdot 29}

u = \frac{- 40 + 4 13}{2 \cdot 29} : \frac{2 ( 13 - 10 )}{29}

\frac{- 40 + 4 13}{2 \cdot 29}

Multiply the numbers:

2 \cdot 29 = 58

= \frac{- 40 + 4 13}{58}

Factor

- 40 + 413 : 4 (- 10 + 13)

- 40 + 413

Rewrite as

= - 4 \cdot 10 + 413

Factor out common term

4

= 4 (- 10 + 13)

= \frac{4 ( - 10 + 13 )}{58}

Cancel the common factor:

2

= \frac{2 ( 13 - 10 )}{29}

u = \frac{- 40 - 4 13}{2 \cdot 29} : - \frac{2 ( 10 + 13 )}{29}

\frac{- 40 - 4 13}{2 \cdot 29}

Multiply the numbers:

2 \cdot 29 = 58

= \frac{- 40 - 4 13}{58}

Factor

- 40 - 413 : - 4 (10 + 13)

- 40 - 413

Rewrite as

= - 4 \cdot 10 - 413

Factor out common term

4

= - 4 (10 + 13)

= - \frac{4 ( 10 + 13 )}{58}

Cancel the common factor:

2

= - \frac{2 ( 10 + 13 )}{29}

The solutions to the quadratic equation are:

u = \frac{2 ( 13 - 10 )}{29}, u = - \frac{2 ( 10 + 13 )}{29}

Substitute back

u = cos (x)

cos (x) = \frac{2 ( 13 - 10 )}{29}, cos (x) = - \frac{2 ( 10 + 13 )}{29}

cos (x) = \frac{2 ( 13 - 10 )}{29}, cos (x) = - \frac{2 ( 10 + 13 )}{29}

cos (x) = \frac{2 ( 13 - 10 )}{29} : x = arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn, x = - arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn

cos (x) = \frac{2 ( 13 - 10 )}{29}

Apply trig inverse properties

cos (x) = \frac{2 ( 13 - 10 )}{29}

General solutions for

cos (x) = \frac{2 ( 13 - 10 )}{29}

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

x = arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn, x = - arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn

x = arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn, x = - arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn

cos (x) = - \frac{2 ( 10 + 13 )}{29} : x = arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn, x = - arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

cos (x) = - \frac{2 ( 10 + 13 )}{29}

Apply trig inverse properties

cos (x) = - \frac{2 ( 10 + 13 )}{29}

General solutions for

cos (x) = - \frac{2 ( 10 + 13 )}{29}

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

x = arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn, x = - arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

x = arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn, x = - arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

Combine all the solutions

x = arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn, x = - arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn, x = arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn, x = - arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

5 cos (x) + 4 - 2 sin (x) = 0

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

Check the solution

arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn :

True

arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn

Plug in

n = 1

arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1

For

5 cos (x) + 4 - 2 sin (x) = 0

plug in

x = arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1

5 cos (arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1) + 4 - 2 sin (arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1) = 0

Refine

0 = 0

\Rightarrow True

Check the solution

- arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn :

False

- arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn

Plug in

n = 1

- arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1

For

5 cos (x) + 4 - 2 sin (x) = 0

plug in

x = - arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1

5 cos (- arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1) + 4 - 2 sin (- arccos (\frac{2 ( 13 - 10 )}{29}) + 2 π 1) = 0

Refine

3.59003 \dots = 0

\Rightarrow False

Check the solution

arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn :

False

arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

Plug in

n = 1

arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1

For

5 cos (x) + 4 - 2 sin (x) = 0

plug in

x = arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1

5 cos (arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1) + 4 - 2 sin (arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1) = 0

Refine

- 1.38313 \dots = 0

\Rightarrow False

Check the solution

- arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn :

True

- arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

Plug in

n = 1

- arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1

For

5 cos (x) + 4 - 2 sin (x) = 0

plug in

x = - arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1

5 cos (- arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1) + 4 - 2 sin (- arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 π 1) = 0

Refine

0 = 0

\Rightarrow True

x = arccos (\frac{2 ( 13 - 10 )}{29}) + 2 πn, x = - arccos (- \frac{2 ( 10 + 13 )}{29}) + 2 πn

Show solutions in decimal form

x = 2.02750 \dots + 2 πn, x = - 2.78851 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 5*cos(x)+4-2*sin(x)=0 ?
The general solution for 5*cos(x)+4-2*sin(x)=0 is x=2.02750…+2pin,x=-2.78851…+2pin