What is the general solution for -15sin(x)-8cos(x)=10 ?

The general solution for -15sin(x)-8cos(x)=10 is x=-3.00267…+2pin,x=2pi-1.11883…+2pin

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Popular Trigonometry >

-15sin(x)-8cos(x)=10

Solution

- 15 sin (x) - 8 cos (x) = 10

Solution

x = - 3.00267 \dots + 2 πn, x = 2 π - 1.11883 \dots + 2 πn

Degrees

x = - 172.04060 \dots^{\circ} + 36 0^{\circ} n, x = 295.89563 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 15 sin (x) - 8 cos (x) = 10

Add

8 cos (x)

to both sides

- 15 sin (x) = 10 + 8 cos (x)

Square both sides

(- 15 sin (x))^{2} = (10 + 8 cos (x))^{2}

Subtract

(10 + 8 cos (x))^{2}

from both sides

225 sin^{2} (x) - 100 - 160 cos (x) - 64 cos^{2} (x) = 0

Rewrite using trig identities

- 100 - 160 cos (x) + 225 sin^{2} (x) - 64 cos^{2} (x)

Use the Pythagorean identity:

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

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

= - 100 - 160 cos (x) + 225 (1 - cos^{2} (x)) - 64 cos^{2} (x)

Simplify

- 100 - 160 cos (x) + 225 (1 - cos^{2} (x)) - 64 cos^{2} (x) : - 289 cos^{2} (x) - 160 cos (x) + 125

- 100 - 160 cos (x) + 225 (1 - cos^{2} (x)) - 64 cos^{2} (x)

Expand

225 (1 - cos^{2} (x)) : 225 - 225 cos^{2} (x)

225 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 225, b = 1, c = cos^{2} (x)

= 225 \cdot 1 - 225 cos^{2} (x)

Multiply the numbers:

225 \cdot 1 = 225

= 225 - 225 cos^{2} (x)

= - 100 - 160 cos (x) + 225 - 225 cos^{2} (x) - 64 cos^{2} (x)

Simplify

- 100 - 160 cos (x) + 225 - 225 cos^{2} (x) - 64 cos^{2} (x) : - 289 cos^{2} (x) - 160 cos (x) + 125

- 100 - 160 cos (x) + 225 - 225 cos^{2} (x) - 64 cos^{2} (x)

Add similar elements:

- 225 cos^{2} (x) - 64 cos^{2} (x) = - 289 cos^{2} (x)

= - 100 - 160 cos (x) + 225 - 289 cos^{2} (x)

Group like terms

= - 160 cos (x) - 289 cos^{2} (x) - 100 + 225

Add/Subtract the numbers:

- 100 + 225 = 125

= - 289 cos^{2} (x) - 160 cos (x) + 125

= - 289 cos^{2} (x) - 160 cos (x) + 125

= - 289 cos^{2} (x) - 160 cos (x) + 125

125 - 160 cos (x) - 289 cos^{2} (x) = 0

Solve by substitution

125 - 160 cos (x) - 289 cos^{2} (x) = 0

Let:

cos (x) = u

125 - 160 u - 289 u^{2} = 0

125 - 160 u - 289 u^{2} = 0 : u = - \frac{5 ( 16 + 9 21 )}{289}, u = \frac{5 ( 9 21 - 16 )}{289}

125 - 160 u - 289 u^{2} = 0

Write in the standard form

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

- 289 u^{2} - 160 u + 125 = 0

Solve with the quadratic formula

- 289 u^{2} - 160 u + 125 = 0

Quadratic Equation Formula:

For

a = - 289, b = - 160, c = 125

u_{1, 2} = \frac{- ( - 160 ) \pm ( - 160 ) ^{2} - 4 ( - 289 ) \cdot 125}{2 ( - 289 )}

u_{1, 2} = \frac{- ( - 160 ) \pm ( - 160 ) ^{2} - 4 ( - 289 ) \cdot 125}{2 ( - 289 )}

(- 160)^{2} - 4 (- 289) \cdot 125 = 9021

(- 160)^{2} - 4 (- 289) \cdot 125

Apply rule

- (- a) = a

= (- 160)^{2} + 4 \cdot 289 \cdot 125

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 160)^{2} = 16 0^{2}

= 16 0^{2} + 4 \cdot 289 \cdot 125

Multiply the numbers:

4 \cdot 289 \cdot 125 = 144500

= 16 0^{2} + 144500

16 0^{2} = 25600

= 25600 + 144500

Add the numbers:

25600 + 144500 = 170100

= 170100

Prime factorization of

170100 : 2^{2} \cdot 3^{5} \cdot 5^{2} \cdot 7

170100

= 3^{5} \cdot 2^{2} \cdot 5^{2} \cdot 7

Apply exponent rule:

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

= 3^{4} \cdot 2^{2} \cdot 5^{2} \cdot 3 \cdot 7

Apply radical rule:

= 2^{2} 3^{4} 5^{2} 3 \cdot 7

Apply radical rule:

2^{2} = 2

= 2 3^{4} 5^{2} 3 \cdot 7

Apply radical rule:

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

= 3^{2} \cdot 2 5^{2} 3 \cdot 7

Apply radical rule:

5^{2} = 5

= 3^{2} \cdot 2 \cdot 5 3 \cdot 7

Refine

= 9021

u_{1, 2} = \frac{- ( - 160 ) \pm 90 21}{2 ( - 289 )}

Separate the solutions

u_{1} = \frac{- ( - 160 ) + 90 21}{2 ( - 289 )}, u_{2} = \frac{- ( - 160 ) - 90 21}{2 ( - 289 )}

u = \frac{- ( - 160 ) + 90 21}{2 ( - 289 )} : - \frac{5 ( 16 + 9 21 )}{289}

\frac{- ( - 160 ) + 90 21}{2 ( - 289 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{160 + 90 21}{- 2 \cdot 289}

Multiply the numbers:

2 \cdot 289 = 578

= \frac{160 + 90 21}{- 578}

Apply the fraction rule:

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

= - \frac{160 + 90 21}{578}

Cancel

\frac{160 + 90 21}{578} : \frac{5 ( 16 + 9 21 )}{289}

\frac{160 + 90 21}{578}

Factor

160 + 9021 : 10 (16 + 921)

160 + 9021

Rewrite as

= 10 \cdot 16 + 10 \cdot 921

Factor out common term

10

= 10 (16 + 921)

= \frac{10 ( 16 + 9 21 )}{578}

Cancel the common factor:

2

= \frac{5 ( 16 + 9 21 )}{289}

= - \frac{5 ( 16 + 9 21 )}{289}

u = \frac{- ( - 160 ) - 90 21}{2 ( - 289 )} : \frac{5 ( 9 21 - 16 )}{289}

\frac{- ( - 160 ) - 90 21}{2 ( - 289 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{160 - 90 21}{- 2 \cdot 289}

Multiply the numbers:

2 \cdot 289 = 578

= \frac{160 - 90 21}{- 578}

Apply the fraction rule:

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

160 - 9021 = - (9021 - 160)

= \frac{90 21 - 160}{578}

Factor

9021 - 160 : 10 (921 - 16)

9021 - 160

Rewrite as

= 10 \cdot 921 - 10 \cdot 16

Factor out common term

10

= 10 (921 - 16)

= \frac{10 ( 9 21 - 16 )}{578}

Cancel the common factor:

2

= \frac{5 ( 9 21 - 16 )}{289}

The solutions to the quadratic equation are:

u = - \frac{5 ( 16 + 9 21 )}{289}, u = \frac{5 ( 9 21 - 16 )}{289}

Substitute back

u = cos (x)

cos (x) = - \frac{5 ( 16 + 9 21 )}{289}, cos (x) = \frac{5 ( 9 21 - 16 )}{289}

cos (x) = - \frac{5 ( 16 + 9 21 )}{289}, cos (x) = \frac{5 ( 9 21 - 16 )}{289}

cos (x) = - \frac{5 ( 16 + 9 21 )}{289} : x = arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 πn, x = - arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 πn

cos (x) = - \frac{5 ( 16 + 9 21 )}{289}

Apply trig inverse properties

cos (x) = - \frac{5 ( 16 + 9 21 )}{289}

General solutions for

cos (x) = - \frac{5 ( 16 + 9 21 )}{289}

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

x = arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 πn, x = - arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 πn

x = arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 πn, x = - arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 πn

cos (x) = \frac{5 ( 9 21 - 16 )}{289} : x = arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 πn, x = 2 π - arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 πn

cos (x) = \frac{5 ( 9 21 - 16 )}{289}

Apply trig inverse properties

cos (x) = \frac{5 ( 9 21 - 16 )}{289}

General solutions for

cos (x) = \frac{5 ( 9 21 - 16 )}{289}

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

x = arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 πn, x = 2 π - arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 πn

x = arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 πn, x = 2 π - arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 πn

Combine all the solutions

x = arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 πn, x = - arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 πn, x = arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 πn, x = 2 π - arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

- 15 sin (x) - 8 cos (x) = 10

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

Check the solution

arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 πn :

False

arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 πn

Plug in

n = 1

arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 π 1

For

- 15 sin (x) - 8 cos (x) = 10

plug in

x = arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 π 1

- 15 sin (arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 π 1) - 8 cos (arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 π 1) = 10

Refine

5.84586 \dots = 10

\Rightarrow False

Check the solution

- arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 πn :

True

- arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 πn

Plug in

n = 1

- arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 π 1

For

- 15 sin (x) - 8 cos (x) = 10

plug in

x = - arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 π 1

- 15 sin (- arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 π 1) - 8 cos (- arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 π 1) = 10

Refine

10 = 10

\Rightarrow True

Check the solution

arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 πn :

False

arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 πn

Plug in

n = 1

arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 π 1

For

- 15 sin (x) - 8 cos (x) = 10

plug in

x = arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 π 1

- 15 sin (arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 π 1) - 8 cos (arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 π 1) = 10

Refine

- 16.98773 \dots = 10

\Rightarrow False

Check the solution

2 π - arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 πn :

True

2 π - arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 π 1

For

- 15 sin (x) - 8 cos (x) = 10

plug in

x = 2 π - arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 π 1

- 15 sin (2 π - arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 π 1) - 8 cos (2 π - arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 π 1) = 10

Refine

10 = 10

\Rightarrow True

x = - arccos (- \frac{5 ( 16 + 9 21 )}{289}) + 2 πn, x = 2 π - arccos (\frac{5 ( 9 21 - 16 )}{289}) + 2 πn

Show solutions in decimal form

x = - 3.00267 \dots + 2 πn, x = 2 π - 1.11883 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for -15sin(x)-8cos(x)=10 ?
The general solution for -15sin(x)-8cos(x)=10 is x=-3.00267…+2pin,x=2pi-1.11883…+2pin