What is the general solution for 9sin(x)+6cos(x)=10 ?

The general solution for 9sin(x)+6cos(x)=10 is x=1.37386…+2pin,x=0.59172…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

9sin(x)+6cos(x)=10

Solution

9 sin (x) + 6 cos (x) = 10

Solution

x = 1.37386 \dots + 2 πn, x = 0.59172 \dots + 2 πn

Degrees

x = 78.71680 \dots^{\circ} + 36 0^{\circ} n, x = 33.90306 \dots^{\circ} + 36 0^{\circ} n

Solution steps

9 sin (x) + 6 cos (x) = 10

Subtract

6 cos (x)

from both sides

9 sin (x) = 10 - 6 cos (x)

Square both sides

(9 sin (x))^{2} = (10 - 6 cos (x))^{2}

Subtract

(10 - 6 cos (x))^{2}

from both sides

81 sin^{2} (x) - 100 + 120 cos (x) - 36 cos^{2} (x) = 0

Rewrite using trig identities

- 100 + 120 cos (x) - 36 cos^{2} (x) + 81 sin^{2} (x)

Use the Pythagorean identity:

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

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

= - 100 + 120 cos (x) - 36 cos^{2} (x) + 81 (1 - cos^{2} (x))

Simplify

- 100 + 120 cos (x) - 36 cos^{2} (x) + 81 (1 - cos^{2} (x)) : 120 cos (x) - 117 cos^{2} (x) - 19

- 100 + 120 cos (x) - 36 cos^{2} (x) + 81 (1 - cos^{2} (x))

Expand

81 (1 - cos^{2} (x)) : 81 - 81 cos^{2} (x)

81 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 81, b = 1, c = cos^{2} (x)

= 81 \cdot 1 - 81 cos^{2} (x)

Multiply the numbers:

81 \cdot 1 = 81

= 81 - 81 cos^{2} (x)

= - 100 + 120 cos (x) - 36 cos^{2} (x) + 81 - 81 cos^{2} (x)

Simplify

- 100 + 120 cos (x) - 36 cos^{2} (x) + 81 - 81 cos^{2} (x) : 120 cos (x) - 117 cos^{2} (x) - 19

- 100 + 120 cos (x) - 36 cos^{2} (x) + 81 - 81 cos^{2} (x)

Group like terms

= 120 cos (x) - 36 cos^{2} (x) - 81 cos^{2} (x) - 100 + 81

Add similar elements:

- 36 cos^{2} (x) - 81 cos^{2} (x) = - 117 cos^{2} (x)

= 120 cos (x) - 117 cos^{2} (x) - 100 + 81

Add/Subtract the numbers:

- 100 + 81 = - 19

= 120 cos (x) - 117 cos^{2} (x) - 19

= 120 cos (x) - 117 cos^{2} (x) - 19

= 120 cos (x) - 117 cos^{2} (x) - 19

- 19 - 117 cos^{2} (x) + 120 cos (x) = 0

Solve by substitution

- 19 - 117 cos^{2} (x) + 120 cos (x) = 0

Let:

cos (x) = u

- 19 - 117 u^{2} + 120 u = 0

- 19 - 117 u^{2} + 120 u = 0 : u = \frac{20 - 3 17}{39}, u = \frac{20 + 3 17}{39}

- 19 - 117 u^{2} + 120 u = 0

Write in the standard form

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

- 117 u^{2} + 120 u - 19 = 0

Solve with the quadratic formula

- 117 u^{2} + 120 u - 19 = 0

Quadratic Equation Formula:

For

a = - 117, b = 120, c = - 19

u_{1, 2} = \frac{- 120 \pm 12 0 ^{2} - 4 ( - 117 ) ( - 19 )}{2 ( - 117 )}

u_{1, 2} = \frac{- 120 \pm 12 0 ^{2} - 4 ( - 117 ) ( - 19 )}{2 ( - 117 )}

12 0^{2} - 4 (- 117) (- 19) = 1817

12 0^{2} - 4 (- 117) (- 19)

Apply rule

- (- a) = a

= 12 0^{2} - 4 \cdot 117 \cdot 19

Multiply the numbers:

4 \cdot 117 \cdot 19 = 8892

= 12 0^{2} - 8892

12 0^{2} = 14400

= 14400 - 8892

Subtract the numbers:

14400 - 8892 = 5508

= 5508

Prime factorization of

5508 : 2^{2} \cdot 3^{4} \cdot 17

5508

5508

divides by

25508 = 2754 \cdot 2

= 2 \cdot 2754

2754

divides by

22754 = 1377 \cdot 2

= 2 \cdot 2 \cdot 1377

1377

divides by

31377 = 459 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 459

459

divides by

3459 = 153 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 3 \cdot 153

153

divides by

3153 = 51 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 51

51

divides by

351 = 17 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 17

2, 3, 17

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 17

= 2^{2} \cdot 3^{4} \cdot 17

= 3^{4} \cdot 2^{2} \cdot 17

Apply radical rule:

n ab = n a n b

= 17 2^{2} 3^{4}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 217 3^{4}

Apply radical rule:

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

3^{4} = 3^{\frac{4}{2}} = 3^{2}

= 3^{2} \cdot 217

Refine

= 1817

u_{1, 2} = \frac{- 120 \pm 18 17}{2 ( - 117 )}

Separate the solutions

u_{1} = \frac{- 120 + 18 17}{2 ( - 117 )}, u_{2} = \frac{- 120 - 18 17}{2 ( - 117 )}

u = \frac{- 120 + 18 17}{2 ( - 117 )} : \frac{20 - 3 17}{39}

\frac{- 120 + 18 17}{2 ( - 117 )}

Remove parentheses:

(- a) = - a

= \frac{- 120 + 18 17}{- 2 \cdot 117}

Multiply the numbers:

2 \cdot 117 = 234

= \frac{- 120 + 18 17}{- 234}

Apply the fraction rule:

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

- 120 + 1817 = - (120 - 1817)

= \frac{120 - 18 17}{234}

Factor

120 - 1817 : 6 (20 - 317)

120 - 1817

Rewrite as

= 6 \cdot 20 - 6 \cdot 317

Factor out common term

6

= 6 (20 - 317)

= \frac{6 ( 20 - 3 17 )}{234}

Cancel the common factor:

6

= \frac{20 - 3 17}{39}

u = \frac{- 120 - 18 17}{2 ( - 117 )} : \frac{20 + 3 17}{39}

\frac{- 120 - 18 17}{2 ( - 117 )}

Remove parentheses:

(- a) = - a

= \frac{- 120 - 18 17}{- 2 \cdot 117}

Multiply the numbers:

2 \cdot 117 = 234

= \frac{- 120 - 18 17}{- 234}

Apply the fraction rule:

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

- 120 - 1817 = - (120 + 1817)

= \frac{120 + 18 17}{234}

Factor

120 + 1817 : 6 (20 + 317)

120 + 1817

Rewrite as

= 6 \cdot 20 + 6 \cdot 317

Factor out common term

6

= 6 (20 + 317)

= \frac{6 ( 20 + 3 17 )}{234}

Cancel the common factor:

6

= \frac{20 + 3 17}{39}

The solutions to the quadratic equation are:

u = \frac{20 - 3 17}{39}, u = \frac{20 + 3 17}{39}

Substitute back

u = cos (x)

cos (x) = \frac{20 - 3 17}{39}, cos (x) = \frac{20 + 3 17}{39}

cos (x) = \frac{20 - 3 17}{39}, cos (x) = \frac{20 + 3 17}{39}

cos (x) = \frac{20 - 3 17}{39} : x = arccos (\frac{20 - 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 - 3 17}{39}) + 2 πn

cos (x) = \frac{20 - 3 17}{39}

Apply trig inverse properties

cos (x) = \frac{20 - 3 17}{39}

General solutions for

cos (x) = \frac{20 - 3 17}{39}

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

x = arccos (\frac{20 - 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 - 3 17}{39}) + 2 πn

x = arccos (\frac{20 - 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 - 3 17}{39}) + 2 πn

cos (x) = \frac{20 + 3 17}{39} : x = arccos (\frac{20 + 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 + 3 17}{39}) + 2 πn

cos (x) = \frac{20 + 3 17}{39}

Apply trig inverse properties

cos (x) = \frac{20 + 3 17}{39}

General solutions for

cos (x) = \frac{20 + 3 17}{39}

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

x = arccos (\frac{20 + 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 + 3 17}{39}) + 2 πn

x = arccos (\frac{20 + 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 + 3 17}{39}) + 2 πn

Combine all the solutions

x = arccos (\frac{20 - 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 - 3 17}{39}) + 2 πn, x = arccos (\frac{20 + 3 17}{39}) + 2 πn, x = 2 π - arccos (\frac{20 + 3 17}{39}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

9 sin (x) + 6 cos (x) = 10

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

Check the solution

arccos (\frac{20 - 3 17}{39}) + 2 πn :

True

arccos (\frac{20 - 3 17}{39}) + 2 πn

Plug in

n = 1

arccos (\frac{20 - 3 17}{39}) + 2 π 1

For

9 sin (x) + 6 cos (x) = 10

plug in

x = arccos (\frac{20 - 3 17}{39}) + 2 π 1

9 sin (arccos (\frac{20 - 3 17}{39}) + 2 π 1) + 6 cos (arccos (\frac{20 - 3 17}{39}) + 2 π 1) = 10

Refine

10 = 10

\Rightarrow True

Check the solution

2 π - arccos (\frac{20 - 3 17}{39}) + 2 πn :

False

2 π - arccos (\frac{20 - 3 17}{39}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{20 - 3 17}{39}) + 2 π 1

For

9 sin (x) + 6 cos (x) = 10

plug in

x = 2 π - arccos (\frac{20 - 3 17}{39}) + 2 π 1

9 sin (2 π - arccos (\frac{20 - 3 17}{39}) + 2 π 1) + 6 cos (2 π - arccos (\frac{20 - 3 17}{39}) + 2 π 1) = 10

Refine

- 7.65209 \dots = 10

\Rightarrow False

Check the solution

arccos (\frac{20 + 3 17}{39}) + 2 πn :

True

arccos (\frac{20 + 3 17}{39}) + 2 πn

Plug in

n = 1

arccos (\frac{20 + 3 17}{39}) + 2 π 1

For

9 sin (x) + 6 cos (x) = 10

plug in

x = arccos (\frac{20 + 3 17}{39}) + 2 π 1

9 sin (arccos (\frac{20 + 3 17}{39}) + 2 π 1) + 6 cos (arccos (\frac{20 + 3 17}{39}) + 2 π 1) = 10

Refine

10 = 10

\Rightarrow True

Check the solution

2 π - arccos (\frac{20 + 3 17}{39}) + 2 πn :

False

2 π - arccos (\frac{20 + 3 17}{39}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{20 + 3 17}{39}) + 2 π 1

For

9 sin (x) + 6 cos (x) = 10

plug in

x = 2 π - arccos (\frac{20 + 3 17}{39}) + 2 π 1

9 sin (2 π - arccos (\frac{20 + 3 17}{39}) + 2 π 1) + 6 cos (2 π - arccos (\frac{20 + 3 17}{39}) + 2 π 1) = 10

Refine

- 0.04021 \dots = 10

\Rightarrow False

x = arccos (\frac{20 - 3 17}{39}) + 2 πn, x = arccos (\frac{20 + 3 17}{39}) + 2 πn

Show solutions in decimal form

x = 1.37386 \dots + 2 πn, x = 0.59172 \dots + 2 πn

Graph

Popular Examples

cos(x)= 4/21

-4csc(x)=-csc^2(x)-4

csc(7x+3)=sqrt(2)

2sin^4(x)-2cos^4(x)=1

18sin(x)=18cos(x)

Frequently Asked Questions (FAQ)

What is the general solution for 9sin(x)+6cos(x)=10 ?
The general solution for 9sin(x)+6cos(x)=10 is x=1.37386…+2pin,x=0.59172…+2pin