What is the general solution for 8sin(θ)+15cos(θ)=18 ?

The general solution for 8sin(θ)+15cos(θ)=18 is No Solution for θ\in\mathbb{R}

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8sin(θ)+15cos(θ)=18

Solution

8 sin (θ) + 15 cos (θ) = 18

Solution

No Solution for θ \in R

Solution steps

8 sin (θ) + 15 cos (θ) = 18

Subtract

15 cos (θ)

from both sides

8 sin (θ) = 18 - 15 cos (θ)

Square both sides

(8 sin (θ))^{2} = (18 - 15 cos (θ))^{2}

Subtract

(18 - 15 cos (θ))^{2}

from both sides

64 sin^{2} (θ) - 324 + 540 cos (θ) - 225 cos^{2} (θ) = 0

Rewrite using trig identities

- 324 - 225 cos^{2} (θ) + 540 cos (θ) + 64 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= - 324 - 225 cos^{2} (θ) + 540 cos (θ) + 64 (1 - cos^{2} (θ))

Simplify

- 324 - 225 cos^{2} (θ) + 540 cos (θ) + 64 (1 - cos^{2} (θ)) : 540 cos (θ) - 289 cos^{2} (θ) - 260

- 324 - 225 cos^{2} (θ) + 540 cos (θ) + 64 (1 - cos^{2} (θ))

Expand

64 (1 - cos^{2} (θ)) : 64 - 64 cos^{2} (θ)

64 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 64, b = 1, c = cos^{2} (θ)

= 64 \cdot 1 - 64 cos^{2} (θ)

Multiply the numbers:

64 \cdot 1 = 64

= 64 - 64 cos^{2} (θ)

= - 324 - 225 cos^{2} (θ) + 540 cos (θ) + 64 - 64 cos^{2} (θ)

Simplify

- 324 - 225 cos^{2} (θ) + 540 cos (θ) + 64 - 64 cos^{2} (θ) : 540 cos (θ) - 289 cos^{2} (θ) - 260

- 324 - 225 cos^{2} (θ) + 540 cos (θ) + 64 - 64 cos^{2} (θ)

Group like terms

= - 225 cos^{2} (θ) + 540 cos (θ) - 64 cos^{2} (θ) - 324 + 64

Add similar elements:

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

= - 289 cos^{2} (θ) + 540 cos (θ) - 324 + 64

Add/Subtract the numbers:

- 324 + 64 = - 260

= 540 cos (θ) - 289 cos^{2} (θ) - 260

= 540 cos (θ) - 289 cos^{2} (θ) - 260

= 540 cos (θ) - 289 cos^{2} (θ) - 260

- 260 - 289 cos^{2} (θ) + 540 cos (θ) = 0

Solve by substitution

- 260 - 289 cos^{2} (θ) + 540 cos (θ) = 0

Let:

cos (θ) = u

- 260 - 289 u^{2} + 540 u = 0

- 260 - 289 u^{2} + 540 u = 0 : u = \frac{270}{289} - i \frac{8 35}{289}, u = \frac{270}{289} + i \frac{8 35}{289}

- 260 - 289 u^{2} + 540 u = 0

Write in the standard form

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

- 289 u^{2} + 540 u - 260 = 0

Solve with the quadratic formula

- 289 u^{2} + 540 u - 260 = 0

Quadratic Equation Formula:

For

a = - 289, b = 540, c = - 260

u_{1, 2} = \frac{- 540 \pm 54 0 ^{2} - 4 ( - 289 ) ( - 260 )}{2 ( - 289 )}

u_{1, 2} = \frac{- 540 \pm 54 0 ^{2} - 4 ( - 289 ) ( - 260 )}{2 ( - 289 )}

Simplify

54 0^{2} - 4 (- 289) (- 260) : 1635 i

54 0^{2} - 4 (- 289) (- 260)

Apply rule

- (- a) = a

= 54 0^{2} - 4 \cdot 289 \cdot 260

Multiply the numbers:

4 \cdot 289 \cdot 260 = 300560

= 54 0^{2} - 300560

Apply imaginary number rule:

- a = i a

= i 300560 - 54 0^{2}

- 54 0^{2} + 300560 = 1635

- 54 0^{2} + 300560

54 0^{2} = 291600

= - 291600 + 300560

Add/Subtract the numbers:

- 291600 + 300560 = 8960

= 8960

Prime factorization of

8960 : 2^{8} \cdot 5 \cdot 7

8960

8960

divides by

28960 = 4480 \cdot 2

= 2 \cdot 4480

4480

divides by

24480 = 2240 \cdot 2

= 2 \cdot 2 \cdot 2240

2240

divides by

22240 = 1120 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 1120

1120

divides by

21120 = 560 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 560

560

divides by

2560 = 280 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 280

280

divides by

2280 = 140 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 140

140

divides by

2140 = 70 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 70

70

divides by

270 = 35 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 35

35

divides by

535 = 7 \cdot 5

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5 \cdot 7

2, 5, 7

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5 \cdot 7

= 2^{8} \cdot 5 \cdot 7

= 2^{8} \cdot 5 \cdot 7

Apply radical rule:

= 2^{8} 5 \cdot 7

Apply radical rule:

2^{8} = 2^{\frac{8}{2}} = 2^{4}

= 2^{4} 5 \cdot 7

Refine

= 1635

= 1635 i

u_{1, 2} = \frac{- 540 \pm 16 35 i}{2 ( - 289 )}

Separate the solutions

u_{1} = \frac{- 540 + 16 35 i}{2 ( - 289 )}, u_{2} = \frac{- 540 - 16 35 i}{2 ( - 289 )}

u = \frac{- 540 + 16 35 i}{2 ( - 289 )} : \frac{270}{289} - i \frac{8 35}{289}

\frac{- 540 + 16 35 i}{2 ( - 289 )}

Remove parentheses:

(- a) = - a

= \frac{- 540 + 16 35 i}{- 2 \cdot 289}

Multiply the numbers:

2 \cdot 289 = 578

= \frac{- 540 + 16 35 i}{- 578}

Apply the fraction rule:

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

= - \frac{- 540 + 16 35 i}{578}

Cancel

\frac{- 540 + 16 35 i}{578} : \frac{2 ( - 135 + 4 35 i )}{289}

\frac{- 540 + 16 35 i}{578}

Factor

- 540 + 1635 i : 4 (- 135 + 435 i)

- 540 + 1635 i

Rewrite as

= - 4 \cdot 135 + 4 \cdot 435 i

Factor out common term

4

= 4 (- 135 + 435 i)

= \frac{4 ( - 135 + 4 35 i )}{578}

Cancel the common factor:

2

= \frac{2 ( - 135 + 4 35 i )}{289}

= - \frac{2 ( - 135 + 4 35 i )}{289}

Rewrite

- \frac{2 ( - 135 + 4 35 i )}{289}

in standard complex form:

\frac{270}{289} - \frac{8 35}{289} i

- \frac{2 ( - 135 + 4 35 i )}{289}

Expand

2 (- 135 + 435 i) : - 270 + 835 i

2 (- 135 + 435 i)

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = - 135, c = 435 i

= 2 (- 135) + 2 \cdot 435 i

Apply minus-plus rules

+ (- a) = - a

= - 2 \cdot 135 + 2 \cdot 435 i

Simplify

- 2 \cdot 135 + 2 \cdot 435 i : - 270 + 835 i

- 2 \cdot 135 + 2 \cdot 435 i

Multiply the numbers:

2 \cdot 135 = 270

= - 270 + 2 \cdot 435 i

Multiply the numbers:

2 \cdot 4 = 8

= - 270 + 835 i

= - 270 + 835 i

= - \frac{- 270 + 8 35 i}{289}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 270 + 8 35 i}{289} = - (- \frac{270}{289}) - (\frac{8 35 i}{289})

= - (- \frac{270}{289}) - (\frac{8 35 i}{289})

Remove parentheses:

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

= \frac{270}{289} - \frac{8 35 i}{289}

= \frac{270}{289} - \frac{8 35}{289} i

u = \frac{- 540 - 16 35 i}{2 ( - 289 )} : \frac{270}{289} + i \frac{8 35}{289}

\frac{- 540 - 16 35 i}{2 ( - 289 )}

Remove parentheses:

(- a) = - a

= \frac{- 540 - 16 35 i}{- 2 \cdot 289}

Multiply the numbers:

2 \cdot 289 = 578

= \frac{- 540 - 16 35 i}{- 578}

Apply the fraction rule:

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

= - \frac{- 540 - 16 35 i}{578}

Cancel

\frac{- 540 - 16 35 i}{578} : - \frac{2 ( 135 + 4 35 i )}{289}

\frac{- 540 - 16 35 i}{578}

Factor

- 540 - 1635 i : - 4 (135 + 435 i)

- 540 - 1635 i

Rewrite as

= - 4 \cdot 135 - 4 \cdot 435 i

Factor out common term

4

= - 4 (135 + 435 i)

= - \frac{4 ( 135 + 4 35 i )}{578}

Cancel the common factor:

2

= - \frac{2 ( 135 + 4 35 i )}{289}

= - (- \frac{2 ( 135 + 4 35 i )}{289})

Apply rule

- (- a) = a

= \frac{2 ( 135 + 4 35 i )}{289}

Rewrite

\frac{2 ( 135 + 4 35 i )}{289}

in standard complex form:

\frac{270}{289} + \frac{8 35}{289} i

\frac{2 ( 135 + 4 35 i )}{289}

Expand

2 (135 + 435 i) : 270 + 835 i

2 (135 + 435 i)

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = 135, c = 435 i

= 2 \cdot 135 + 2 \cdot 435 i

Simplify

2 \cdot 135 + 2 \cdot 435 i : 270 + 835 i

2 \cdot 135 + 2 \cdot 435 i

Multiply the numbers:

2 \cdot 135 = 270

= 270 + 2 \cdot 435 i

Multiply the numbers:

2 \cdot 4 = 8

= 270 + 835 i

= 270 + 835 i

= \frac{270 + 8 35 i}{289}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{270 + 8 35 i}{289} = \frac{270}{289} + \frac{8 35 i}{289}

= \frac{270}{289} + \frac{8 35 i}{289}

= \frac{270}{289} + \frac{8 35}{289} i

The solutions to the quadratic equation are:

u = \frac{270}{289} - i \frac{8 35}{289}, u = \frac{270}{289} + i \frac{8 35}{289}

Substitute back

u = cos (θ)

cos (θ) = \frac{270}{289} - i \frac{8 35}{289}, cos (θ) = \frac{270}{289} + i \frac{8 35}{289}

cos (θ) = \frac{270}{289} - i \frac{8 35}{289}, cos (θ) = \frac{270}{289} + i \frac{8 35}{289}

cos (θ) = \frac{270}{289} - i \frac{8 35}{289} :

No Solution

cos (θ) = \frac{270}{289} - i \frac{8 35}{289}

No Solution

cos (θ) = \frac{270}{289} + i \frac{8 35}{289} :

No Solution

cos (θ) = \frac{270}{289} + i \frac{8 35}{289}

No Solution

Combine all the solutions

No Solution

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

8 sin (θ) + 15 cos (θ) = 18

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

No Solution for θ \in R

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Frequently Asked Questions (FAQ)

What is the general solution for 8sin(θ)+15cos(θ)=18 ?
The general solution for 8sin(θ)+15cos(θ)=18 is No Solution for θ\in\mathbb{R}