What is the general solution for 5sin(θ)+12cos(θ)=13 ?

The general solution for 5sin(θ)+12cos(θ)=13 is θ=0.39479…+2pin

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Popular >

5sin(θ)+12cos(θ)=13

Solution

5 sin (θ) + 12 cos (θ) = 13

Solution

θ = 0.39479 \dots + 2 πn

Degrees

θ = 22.61986 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 sin (θ) + 12 cos (θ) = 13

Subtract

12 cos (θ)

from both sides

5 sin (θ) = 13 - 12 cos (θ)

Square both sides

(5 sin (θ))^{2} = (13 - 12 cos (θ))^{2}

Subtract

(13 - 12 cos (θ))^{2}

from both sides

25 sin^{2} (θ) - 169 + 312 cos (θ) - 144 cos^{2} (θ) = 0

Rewrite using trig identities

- 169 - 144 cos^{2} (θ) + 25 sin^{2} (θ) + 312 cos (θ)

Use the Pythagorean identity:

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

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

= - 169 - 144 cos^{2} (θ) + 25 (1 - cos^{2} (θ)) + 312 cos (θ)

Simplify

- 169 - 144 cos^{2} (θ) + 25 (1 - cos^{2} (θ)) + 312 cos (θ) : 312 cos (θ) - 169 cos^{2} (θ) - 144

- 169 - 144 cos^{2} (θ) + 25 (1 - cos^{2} (θ)) + 312 cos (θ)

Expand

25 (1 - cos^{2} (θ)) : 25 - 25 cos^{2} (θ)

25 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 25, b = 1, c = cos^{2} (θ)

= 25 \cdot 1 - 25 cos^{2} (θ)

Multiply the numbers:

25 \cdot 1 = 25

= 25 - 25 cos^{2} (θ)

= - 169 - 144 cos^{2} (θ) + 25 - 25 cos^{2} (θ) + 312 cos (θ)

Simplify

- 169 - 144 cos^{2} (θ) + 25 - 25 cos^{2} (θ) + 312 cos (θ) : 312 cos (θ) - 169 cos^{2} (θ) - 144

- 169 - 144 cos^{2} (θ) + 25 - 25 cos^{2} (θ) + 312 cos (θ)

Group like terms

= - 144 cos^{2} (θ) - 25 cos^{2} (θ) + 312 cos (θ) - 169 + 25

Add similar elements:

- 144 cos^{2} (θ) - 25 cos^{2} (θ) = - 169 cos^{2} (θ)

= - 169 cos^{2} (θ) + 312 cos (θ) - 169 + 25

Add/Subtract the numbers:

- 169 + 25 = - 144

= 312 cos (θ) - 169 cos^{2} (θ) - 144

= 312 cos (θ) - 169 cos^{2} (θ) - 144

= 312 cos (θ) - 169 cos^{2} (θ) - 144

- 144 - 169 cos^{2} (θ) + 312 cos (θ) = 0

Solve by substitution

- 144 - 169 cos^{2} (θ) + 312 cos (θ) = 0

Let:

cos (θ) = u

- 144 - 169 u^{2} + 312 u = 0

- 144 - 169 u^{2} + 312 u = 0 : u = \frac{12}{13}

- 144 - 169 u^{2} + 312 u = 0

Write in the standard form

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

- 169 u^{2} + 312 u - 144 = 0

Solve with the quadratic formula

- 169 u^{2} + 312 u - 144 = 0

Quadratic Equation Formula:

For

a = - 169, b = 312, c = - 144

u_{1, 2} = \frac{- 312 \pm 31 2 ^{2} - 4 ( - 169 ) ( - 144 )}{2 ( - 169 )}

u_{1, 2} = \frac{- 312 \pm 31 2 ^{2} - 4 ( - 169 ) ( - 144 )}{2 ( - 169 )}

31 2^{2} - 4 (- 169) (- 144) = 0

31 2^{2} - 4 (- 169) (- 144)

Apply rule

- (- a) = a

= 31 2^{2} - 4 \cdot 169 \cdot 144

Multiply the numbers:

4 \cdot 169 \cdot 144 = 97344

= 31 2^{2} - 97344

31 2^{2} = 97344

= 97344 - 97344

Subtract the numbers:

97344 - 97344 = 0

= 0

u_{1, 2} = \frac{- 312 \pm 0}{2 ( - 169 )}

u = \frac{- 312}{2 ( - 169 )}

\frac{- 312}{2 ( - 169 )} = \frac{12}{13}

\frac{- 312}{2 ( - 169 )}

Remove parentheses:

(- a) = - a

= \frac{- 312}{- 2 \cdot 169}

Multiply the numbers:

2 \cdot 169 = 338

= \frac{- 312}{- 338}

Apply the fraction rule:

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

= \frac{312}{338}

Cancel the common factor:

26

= \frac{12}{13}

u = \frac{12}{13}

The solution to the quadratic equation is:

u = \frac{12}{13}

Substitute back

u = cos (θ)

cos (θ) = \frac{12}{13}

cos (θ) = \frac{12}{13}

cos (θ) = \frac{12}{13} : θ = arccos (\frac{12}{13}) + 2 πn, θ = 2 π - arccos (\frac{12}{13}) + 2 πn

cos (θ) = \frac{12}{13}

Apply trig inverse properties

cos (θ) = \frac{12}{13}

General solutions for

cos (θ) = \frac{12}{13}

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

θ = arccos (\frac{12}{13}) + 2 πn, θ = 2 π - arccos (\frac{12}{13}) + 2 πn

θ = arccos (\frac{12}{13}) + 2 πn, θ = 2 π - arccos (\frac{12}{13}) + 2 πn

Combine all the solutions

θ = arccos (\frac{12}{13}) + 2 πn, θ = 2 π - arccos (\frac{12}{13}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

5 sin (θ) + 12 cos (θ) = 13

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

Check the solution

arccos (\frac{12}{13}) + 2 πn :

True

arccos (\frac{12}{13}) + 2 πn

Plug in

n = 1

arccos (\frac{12}{13}) + 2 π 1

For

5 sin (θ) + 12 cos (θ) = 13

plug in

θ = arccos (\frac{12}{13}) + 2 π 1

5 sin (arccos (\frac{12}{13}) + 2 π 1) + 12 cos (arccos (\frac{12}{13}) + 2 π 1) = 13

Refine

13 = 13

\Rightarrow True

Check the solution

2 π - arccos (\frac{12}{13}) + 2 πn :

False

2 π - arccos (\frac{12}{13}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{12}{13}) + 2 π 1

For

5 sin (θ) + 12 cos (θ) = 13

plug in

θ = 2 π - arccos (\frac{12}{13}) + 2 π 1

5 sin (2 π - arccos (\frac{12}{13}) + 2 π 1) + 12 cos (2 π - arccos (\frac{12}{13}) + 2 π 1) = 13

Refine

9.15384 \dots = 13

\Rightarrow False

θ = arccos (\frac{12}{13}) + 2 πn

Show solutions in decimal form

θ = 0.39479 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5sin(θ)+12cos(θ)=13 ?
The general solution for 5sin(θ)+12cos(θ)=13 is θ=0.39479…+2pin