What is the general solution for 4sin(x)+5cos(x)=6 ?

The general solution for 4sin(x)+5cos(x)=6 is x=1.03147…+2pin,x=0.31800…+2pin

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Popular Trigonometry >

4sin(x)+5cos(x)=6

Solution

4 sin (x) + 5 cos (x) = 6

Solution

x = 1.03147 \dots + 2 πn, x = 0.31800 \dots + 2 πn

Degrees

x = 59.09912 \dots^{\circ} + 36 0^{\circ} n, x = 18.22049 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 sin (x) + 5 cos (x) = 6

Subtract

5 cos (x)

from both sides

4 sin (x) = 6 - 5 cos (x)

Square both sides

(4 sin (x))^{2} = (6 - 5 cos (x))^{2}

Subtract

(6 - 5 cos (x))^{2}

from both sides

16 sin^{2} (x) - 36 + 60 cos (x) - 25 cos^{2} (x) = 0

Rewrite using trig identities

- 36 + 16 sin^{2} (x) - 25 cos^{2} (x) + 60 cos (x)

Use the Pythagorean identity:

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

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

= - 36 + 16 (1 - cos^{2} (x)) - 25 cos^{2} (x) + 60 cos (x)

Simplify

- 36 + 16 (1 - cos^{2} (x)) - 25 cos^{2} (x) + 60 cos (x) : 60 cos (x) - 41 cos^{2} (x) - 20

- 36 + 16 (1 - cos^{2} (x)) - 25 cos^{2} (x) + 60 cos (x)

Expand

16 (1 - cos^{2} (x)) : 16 - 16 cos^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

16 \cdot 1 = 16

= 16 - 16 cos^{2} (x)

= - 36 + 16 - 16 cos^{2} (x) - 25 cos^{2} (x) + 60 cos (x)

Simplify

- 36 + 16 - 16 cos^{2} (x) - 25 cos^{2} (x) + 60 cos (x) : 60 cos (x) - 41 cos^{2} (x) - 20

- 36 + 16 - 16 cos^{2} (x) - 25 cos^{2} (x) + 60 cos (x)

Add similar elements:

- 16 cos^{2} (x) - 25 cos^{2} (x) = - 41 cos^{2} (x)

= - 36 + 16 - 41 cos^{2} (x) + 60 cos (x)

Add/Subtract the numbers:

- 36 + 16 = - 20

= 60 cos (x) - 41 cos^{2} (x) - 20

= 60 cos (x) - 41 cos^{2} (x) - 20

= 60 cos (x) - 41 cos^{2} (x) - 20

- 20 - 41 cos^{2} (x) + 60 cos (x) = 0

Solve by substitution

- 20 - 41 cos^{2} (x) + 60 cos (x) = 0

Let:

cos (x) = u

- 20 - 41 u^{2} + 60 u = 0

- 20 - 41 u^{2} + 60 u = 0 : u = \frac{2 ( 15 - 2 5 )}{41}, u = \frac{2 ( 15 + 2 5 )}{41}

- 20 - 41 u^{2} + 60 u = 0

Write in the standard form

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

- 41 u^{2} + 60 u - 20 = 0

Solve with the quadratic formula

- 41 u^{2} + 60 u - 20 = 0

Quadratic Equation Formula:

For

a = - 41, b = 60, c = - 20

u_{1, 2} = \frac{- 60 \pm 6 0 ^{2} - 4 ( - 41 ) ( - 20 )}{2 ( - 41 )}

u_{1, 2} = \frac{- 60 \pm 6 0 ^{2} - 4 ( - 41 ) ( - 20 )}{2 ( - 41 )}

6 0^{2} - 4 (- 41) (- 20) = 85

6 0^{2} - 4 (- 41) (- 20)

Apply rule

- (- a) = a

= 6 0^{2} - 4 \cdot 41 \cdot 20

Multiply the numbers:

4 \cdot 41 \cdot 20 = 3280

= 6 0^{2} - 3280

6 0^{2} = 3600

= 3600 - 3280

Subtract the numbers:

3600 - 3280 = 320

= 320

Prime factorization of

320 : 2^{6} \cdot 5

320

320

divides by

2320 = 160 \cdot 2

= 2 \cdot 160

160

divides by

2160 = 80 \cdot 2

= 2 \cdot 2 \cdot 80

80

divides by

280 = 40 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 40

40

divides by

240 = 20 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 10

10

divides by

210 = 5 \cdot 2

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

2, 5

are all prime numbers, therefore no further factorization is possible

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

= 2^{6} \cdot 5

= 2^{6} \cdot 5

Apply radical rule:

= 5 2^{6}

Apply radical rule:

2^{6} = 2^{\frac{6}{2}} = 2^{3}

= 2^{3} 5

Refine

= 85

u_{1, 2} = \frac{- 60 \pm 8 5}{2 ( - 41 )}

Separate the solutions

u_{1} = \frac{- 60 + 8 5}{2 ( - 41 )}, u_{2} = \frac{- 60 - 8 5}{2 ( - 41 )}

u = \frac{- 60 + 8 5}{2 ( - 41 )} : \frac{2 ( 15 - 2 5 )}{41}

\frac{- 60 + 8 5}{2 ( - 41 )}

Remove parentheses:

(- a) = - a

= \frac{- 60 + 8 5}{- 2 \cdot 41}

Multiply the numbers:

2 \cdot 41 = 82

= \frac{- 60 + 8 5}{- 82}

Apply the fraction rule:

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

- 60 + 85 = - (60 - 85)

= \frac{60 - 8 5}{82}

Factor

60 - 85 : 4 (15 - 25)

60 - 85

Rewrite as

= 4 \cdot 15 - 4 \cdot 25

Factor out common term

4

= 4 (15 - 25)

= \frac{4 ( 15 - 2 5 )}{82}

Cancel the common factor:

2

= \frac{2 ( 15 - 2 5 )}{41}

u = \frac{- 60 - 8 5}{2 ( - 41 )} : \frac{2 ( 15 + 2 5 )}{41}

\frac{- 60 - 8 5}{2 ( - 41 )}

Remove parentheses:

(- a) = - a

= \frac{- 60 - 8 5}{- 2 \cdot 41}

Multiply the numbers:

2 \cdot 41 = 82

= \frac{- 60 - 8 5}{- 82}

Apply the fraction rule:

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

- 60 - 85 = - (60 + 85)

= \frac{60 + 8 5}{82}

Factor

60 + 85 : 4 (15 + 25)

60 + 85

Rewrite as

= 4 \cdot 15 + 4 \cdot 25

Factor out common term

4

= 4 (15 + 25)

= \frac{4 ( 15 + 2 5 )}{82}

Cancel the common factor:

2

= \frac{2 ( 15 + 2 5 )}{41}

The solutions to the quadratic equation are:

u = \frac{2 ( 15 - 2 5 )}{41}, u = \frac{2 ( 15 + 2 5 )}{41}

Substitute back

u = cos (x)

cos (x) = \frac{2 ( 15 - 2 5 )}{41}, cos (x) = \frac{2 ( 15 + 2 5 )}{41}

cos (x) = \frac{2 ( 15 - 2 5 )}{41}, cos (x) = \frac{2 ( 15 + 2 5 )}{41}

cos (x) = \frac{2 ( 15 - 2 5 )}{41} : x = arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn

cos (x) = \frac{2 ( 15 - 2 5 )}{41}

Apply trig inverse properties

cos (x) = \frac{2 ( 15 - 2 5 )}{41}

General solutions for

cos (x) = \frac{2 ( 15 - 2 5 )}{41}

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

x = arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn

x = arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn

cos (x) = \frac{2 ( 15 + 2 5 )}{41} : x = arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

cos (x) = \frac{2 ( 15 + 2 5 )}{41}

Apply trig inverse properties

cos (x) = \frac{2 ( 15 + 2 5 )}{41}

General solutions for

cos (x) = \frac{2 ( 15 + 2 5 )}{41}

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

x = arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

x = arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

Combine all the solutions

x = arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn, x = arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn, x = 2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

4 sin (x) + 5 cos (x) = 6

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

Check the solution

arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn :

True

arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn

Plug in

n = 1

arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1

For

4 sin (x) + 5 cos (x) = 6

plug in

x = arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1

4 sin (arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1) + 5 cos (arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1) = 6

Refine

6 = 6

\Rightarrow True

Check the solution

2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn :

False

2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1

For

4 sin (x) + 5 cos (x) = 6

plug in

x = 2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1

4 sin (2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1) + 5 cos (2 π - arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 π 1) = 6

Refine

- 0.86445 \dots = 6

\Rightarrow False

Check the solution

arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn :

True

arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

Plug in

n = 1

arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1

For

4 sin (x) + 5 cos (x) = 6

plug in

x = arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1

4 sin (arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1) + 5 cos (arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1) = 6

Refine

6 = 6

\Rightarrow True

Check the solution

2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn :

False

2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1

For

4 sin (x) + 5 cos (x) = 6

plug in

x = 2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1

4 sin (2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1) + 5 cos (2 π - arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 π 1) = 6

Refine

3.49860 \dots = 6

\Rightarrow False

x = arccos (\frac{2 ( 15 - 2 5 )}{41}) + 2 πn, x = arccos (\frac{2 ( 15 + 2 5 )}{41}) + 2 πn

Show solutions in decimal form

x = 1.03147 \dots + 2 πn, x = 0.31800 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin(x)+5cos(x)=6 ?
The general solution for 4sin(x)+5cos(x)=6 is x=1.03147…+2pin,x=0.31800…+2pin