What is the general solution for 8sin(θ)+15cos(θ)=17 ?

The general solution for 8sin(θ)+15cos(θ)=17 is θ=0.48995…+2pin

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Popular Trigonometry >

8sin(θ)+15cos(θ)=17

Solution

8 sin (θ) + 15 cos (θ) = 17

Solution

θ = 0.48995 \dots + 2 πn

Degrees

θ = 28.07248 \dots^{\circ} + 36 0^{\circ} n

Solution steps

8 sin (θ) + 15 cos (θ) = 17

Subtract

15 cos (θ)

from both sides

8 sin (θ) = 17 - 15 cos (θ)

Square both sides

(8 sin (θ))^{2} = (17 - 15 cos (θ))^{2}

Subtract

(17 - 15 cos (θ))^{2}

from both sides

64 sin^{2} (θ) - 289 + 510 cos (θ) - 225 cos^{2} (θ) = 0

Rewrite using trig identities

- 289 - 225 cos^{2} (θ) + 510 cos (θ) + 64 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= - 289 - 225 cos^{2} (θ) + 510 cos (θ) + 64 (1 - cos^{2} (θ))

Simplify

- 289 - 225 cos^{2} (θ) + 510 cos (θ) + 64 (1 - cos^{2} (θ)) : 510 cos (θ) - 289 cos^{2} (θ) - 225

- 289 - 225 cos^{2} (θ) + 510 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)

= - 289 - 225 cos^{2} (θ) + 510 cos (θ) + 64 - 64 cos^{2} (θ)

Simplify

- 289 - 225 cos^{2} (θ) + 510 cos (θ) + 64 - 64 cos^{2} (θ) : 510 cos (θ) - 289 cos^{2} (θ) - 225

- 289 - 225 cos^{2} (θ) + 510 cos (θ) + 64 - 64 cos^{2} (θ)

Group like terms

= - 225 cos^{2} (θ) + 510 cos (θ) - 64 cos^{2} (θ) - 289 + 64

Add similar elements:

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

= - 289 cos^{2} (θ) + 510 cos (θ) - 289 + 64

Add/Subtract the numbers:

- 289 + 64 = - 225

= 510 cos (θ) - 289 cos^{2} (θ) - 225

= 510 cos (θ) - 289 cos^{2} (θ) - 225

= 510 cos (θ) - 289 cos^{2} (θ) - 225

- 225 - 289 cos^{2} (θ) + 510 cos (θ) = 0

Solve by substitution

- 225 - 289 cos^{2} (θ) + 510 cos (θ) = 0

Let:

cos (θ) = u

- 225 - 289 u^{2} + 510 u = 0

- 225 - 289 u^{2} + 510 u = 0 : u = \frac{15}{17}

- 225 - 289 u^{2} + 510 u = 0

Write in the standard form

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

- 289 u^{2} + 510 u - 225 = 0

Solve with the quadratic formula

- 289 u^{2} + 510 u - 225 = 0

Quadratic Equation Formula:

For

a = - 289, b = 510, c = - 225

u_{1, 2} = \frac{- 510 \pm 51 0 ^{2} - 4 ( - 289 ) ( - 225 )}{2 ( - 289 )}

u_{1, 2} = \frac{- 510 \pm 51 0 ^{2} - 4 ( - 289 ) ( - 225 )}{2 ( - 289 )}

51 0^{2} - 4 (- 289) (- 225) = 0

51 0^{2} - 4 (- 289) (- 225)

Apply rule

- (- a) = a

= 51 0^{2} - 4 \cdot 289 \cdot 225

Multiply the numbers:

4 \cdot 289 \cdot 225 = 260100

= 51 0^{2} - 260100

51 0^{2} = 260100

= 260100 - 260100

Subtract the numbers:

260100 - 260100 = 0

= 0

u_{1, 2} = \frac{- 510 \pm 0}{2 ( - 289 )}

u = \frac{- 510}{2 ( - 289 )}

\frac{- 510}{2 ( - 289 )} = \frac{15}{17}

\frac{- 510}{2 ( - 289 )}

Remove parentheses:

(- a) = - a

= \frac{- 510}{- 2 \cdot 289}

Multiply the numbers:

2 \cdot 289 = 578

= \frac{- 510}{- 578}

Apply the fraction rule:

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

= \frac{510}{578}

Cancel the common factor:

34

= \frac{15}{17}

u = \frac{15}{17}

The solution to the quadratic equation is:

u = \frac{15}{17}

Substitute back

u = cos (θ)

cos (θ) = \frac{15}{17}

cos (θ) = \frac{15}{17}

cos (θ) = \frac{15}{17} : θ = arccos (\frac{15}{17}) + 2 πn, θ = 2 π - arccos (\frac{15}{17}) + 2 πn

cos (θ) = \frac{15}{17}

Apply trig inverse properties

cos (θ) = \frac{15}{17}

General solutions for

cos (θ) = \frac{15}{17}

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

θ = arccos (\frac{15}{17}) + 2 πn, θ = 2 π - arccos (\frac{15}{17}) + 2 πn

θ = arccos (\frac{15}{17}) + 2 πn, θ = 2 π - arccos (\frac{15}{17}) + 2 πn

Combine all the solutions

θ = arccos (\frac{15}{17}) + 2 πn, θ = 2 π - arccos (\frac{15}{17}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

8 sin (θ) + 15 cos (θ) = 17

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

Check the solution

arccos (\frac{15}{17}) + 2 πn :

True

arccos (\frac{15}{17}) + 2 πn

Plug in

n = 1

arccos (\frac{15}{17}) + 2 π 1

For

8 sin (θ) + 15 cos (θ) = 17

plug in

θ = arccos (\frac{15}{17}) + 2 π 1

8 sin (arccos (\frac{15}{17}) + 2 π 1) + 15 cos (arccos (\frac{15}{17}) + 2 π 1) = 17

Refine

17 = 17

\Rightarrow True

Check the solution

2 π - arccos (\frac{15}{17}) + 2 πn :

False

2 π - arccos (\frac{15}{17}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{15}{17}) + 2 π 1

For

8 sin (θ) + 15 cos (θ) = 17

plug in

θ = 2 π - arccos (\frac{15}{17}) + 2 π 1

8 sin (2 π - arccos (\frac{15}{17}) + 2 π 1) + 15 cos (2 π - arccos (\frac{15}{17}) + 2 π 1) = 17

Refine

9.47058 \dots = 17

\Rightarrow False

θ = arccos (\frac{15}{17}) + 2 πn

Show solutions in decimal form

θ = 0.48995 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 8sin(θ)+15cos(θ)=17 ?
The general solution for 8sin(θ)+15cos(θ)=17 is θ=0.48995…+2pin