What is the general solution for 5sin(θ)-5cos(θ)=2 ?

The general solution for 5sin(θ)-5cos(θ)=2 is θ=-2.64295…+2pin,θ=1.07215…+2pin

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Popular Trigonometry >

5sin(θ)-5cos(θ)=2

Solution

5 sin (θ) - 5 cos (θ) = 2

Solution

θ = - 2.64295 \dots + 2 πn, θ = 1.07215 \dots + 2 πn

Degrees

θ = - 151.42994 \dots^{\circ} + 36 0^{\circ} n, θ = 61.42994 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 sin (θ) - 5 cos (θ) = 2

Add

5 cos (θ)

to both sides

5 sin (θ) = 2 + 5 cos (θ)

Square both sides

(5 sin (θ))^{2} = (2 + 5 cos (θ))^{2}

Subtract

(2 + 5 cos (θ))^{2}

from both sides

25 sin^{2} (θ) - 4 - 20 cos (θ) - 25 cos^{2} (θ) = 0

Rewrite using trig identities

- 4 - 20 cos (θ) - 25 cos^{2} (θ) + 25 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= - 4 - 20 cos (θ) - 25 cos^{2} (θ) + 25 (1 - cos^{2} (θ))

Simplify

- 4 - 20 cos (θ) - 25 cos^{2} (θ) + 25 (1 - cos^{2} (θ)) : - 50 cos^{2} (θ) - 20 cos (θ) + 21

- 4 - 20 cos (θ) - 25 cos^{2} (θ) + 25 (1 - cos^{2} (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)

= - 4 - 20 cos (θ) - 25 cos^{2} (θ) + 25 - 25 cos^{2} (θ)

Simplify

- 4 - 20 cos (θ) - 25 cos^{2} (θ) + 25 - 25 cos^{2} (θ) : - 50 cos^{2} (θ) - 20 cos (θ) + 21

- 4 - 20 cos (θ) - 25 cos^{2} (θ) + 25 - 25 cos^{2} (θ)

Group like terms

= - 20 cos (θ) - 25 cos^{2} (θ) - 25 cos^{2} (θ) - 4 + 25

Add similar elements:

- 25 cos^{2} (θ) - 25 cos^{2} (θ) = - 50 cos^{2} (θ)

= - 20 cos (θ) - 50 cos^{2} (θ) - 4 + 25

Add/Subtract the numbers:

- 4 + 25 = 21

= - 50 cos^{2} (θ) - 20 cos (θ) + 21

= - 50 cos^{2} (θ) - 20 cos (θ) + 21

= - 50 cos^{2} (θ) - 20 cos (θ) + 21

21 - 20 cos (θ) - 50 cos^{2} (θ) = 0

Solve by substitution

21 - 20 cos (θ) - 50 cos^{2} (θ) = 0

Let:

cos (θ) = u

21 - 20 u - 50 u^{2} = 0

21 - 20 u - 50 u^{2} = 0 : u = - \frac{2 + 46}{10}, u = \frac{46 - 2}{10}

21 - 20 u - 50 u^{2} = 0

Write in the standard form

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

- 50 u^{2} - 20 u + 21 = 0

Solve with the quadratic formula

- 50 u^{2} - 20 u + 21 = 0

Quadratic Equation Formula:

For

a = - 50, b = - 20, c = 21

u_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 ( - 50 ) \cdot 21}{2 ( - 50 )}

u_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 ( - 50 ) \cdot 21}{2 ( - 50 )}

(- 20)^{2} - 4 (- 50) \cdot 21 = 1046

(- 20)^{2} - 4 (- 50) \cdot 21

Apply rule

- (- a) = a

= (- 20)^{2} + 4 \cdot 50 \cdot 21

Apply exponent rule:

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

n

is even

(- 20)^{2} = 2 0^{2}

= 2 0^{2} + 4 \cdot 50 \cdot 21

Multiply the numbers:

4 \cdot 50 \cdot 21 = 4200

= 2 0^{2} + 4200

2 0^{2} = 400

= 400 + 4200

Add the numbers:

400 + 4200 = 4600

= 4600

Prime factorization of

4600 : 2^{3} \cdot 5^{2} \cdot 23

4600

4600

divides by

24600 = 2300 \cdot 2

= 2 \cdot 2300

2300

divides by

22300 = 1150 \cdot 2

= 2 \cdot 2 \cdot 1150

1150

divides by

21150 = 575 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 575

575

divides by

5575 = 115 \cdot 5

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 115

115

divides by

5115 = 23 \cdot 5

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5 \cdot 23

2, 5, 23

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5 \cdot 23

= 2^{3} \cdot 5^{2} \cdot 23

= 2^{3} \cdot 5^{2} \cdot 23

Apply exponent rule:

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

= 2^{2} \cdot 5^{2} \cdot 2 \cdot 23

Apply radical rule:

= 2^{2} 5^{2} 2 \cdot 23

Apply radical rule:

2^{2} = 2

= 2 5^{2} 2 \cdot 23

Apply radical rule:

5^{2} = 5

= 2 \cdot 5 2 \cdot 23

Refine

= 1046

u_{1, 2} = \frac{- ( - 20 ) \pm 10 46}{2 ( - 50 )}

Separate the solutions

u_{1} = \frac{- ( - 20 ) + 10 46}{2 ( - 50 )}, u_{2} = \frac{- ( - 20 ) - 10 46}{2 ( - 50 )}

u = \frac{- ( - 20 ) + 10 46}{2 ( - 50 )} : - \frac{2 + 46}{10}

\frac{- ( - 20 ) + 10 46}{2 ( - 50 )}

Remove parentheses:

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

= \frac{20 + 10 46}{- 2 \cdot 50}

Multiply the numbers:

2 \cdot 50 = 100

= \frac{20 + 10 46}{- 100}

Apply the fraction rule:

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

= - \frac{20 + 10 46}{100}

Cancel

\frac{20 + 10 46}{100} : \frac{2 + 46}{10}

\frac{20 + 10 46}{100}

Factor

20 + 1046 : 10 (2 + 46)

20 + 1046

Rewrite as

= 10 \cdot 2 + 1046

Factor out common term

10

= 10 (2 + 46)

= \frac{10 ( 2 + 46 )}{100}

Cancel the common factor:

10

= \frac{2 + 46}{10}

= - \frac{2 + 46}{10}

u = \frac{- ( - 20 ) - 10 46}{2 ( - 50 )} : \frac{46 - 2}{10}

\frac{- ( - 20 ) - 10 46}{2 ( - 50 )}

Remove parentheses:

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

= \frac{20 - 10 46}{- 2 \cdot 50}

Multiply the numbers:

2 \cdot 50 = 100

= \frac{20 - 10 46}{- 100}

Apply the fraction rule:

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

20 - 1046 = - (1046 - 20)

= \frac{10 46 - 20}{100}

Factor

1046 - 20 : 10 (46 - 2)

1046 - 20

Rewrite as

= 1046 - 10 \cdot 2

Factor out common term

10

= 10 (46 - 2)

= \frac{10 ( 46 - 2 )}{100}

Cancel the common factor:

10

= \frac{46 - 2}{10}

The solutions to the quadratic equation are:

u = - \frac{2 + 46}{10}, u = \frac{46 - 2}{10}

Substitute back

u = cos (θ)

cos (θ) = - \frac{2 + 46}{10}, cos (θ) = \frac{46 - 2}{10}

cos (θ) = - \frac{2 + 46}{10}, cos (θ) = \frac{46 - 2}{10}

cos (θ) = - \frac{2 + 46}{10} : θ = arccos (- \frac{2 + 46}{10}) + 2 πn, θ = - arccos (- \frac{2 + 46}{10}) + 2 πn

cos (θ) = - \frac{2 + 46}{10}

Apply trig inverse properties

cos (θ) = - \frac{2 + 46}{10}

General solutions for

cos (θ) = - \frac{2 + 46}{10}

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

θ = arccos (- \frac{2 + 46}{10}) + 2 πn, θ = - arccos (- \frac{2 + 46}{10}) + 2 πn

θ = arccos (- \frac{2 + 46}{10}) + 2 πn, θ = - arccos (- \frac{2 + 46}{10}) + 2 πn

cos (θ) = \frac{46 - 2}{10} : θ = arccos (\frac{46 - 2}{10}) + 2 πn, θ = 2 π - arccos (\frac{46 - 2}{10}) + 2 πn

cos (θ) = \frac{46 - 2}{10}

Apply trig inverse properties

cos (θ) = \frac{46 - 2}{10}

General solutions for

cos (θ) = \frac{46 - 2}{10}

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

θ = arccos (\frac{46 - 2}{10}) + 2 πn, θ = 2 π - arccos (\frac{46 - 2}{10}) + 2 πn

θ = arccos (\frac{46 - 2}{10}) + 2 πn, θ = 2 π - arccos (\frac{46 - 2}{10}) + 2 πn

Combine all the solutions

θ = arccos (- \frac{2 + 46}{10}) + 2 πn, θ = - arccos (- \frac{2 + 46}{10}) + 2 πn, θ = arccos (\frac{46 - 2}{10}) + 2 πn, θ = 2 π - arccos (\frac{46 - 2}{10}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

5 sin (θ) - 5 cos (θ) = 2

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

Check the solution

arccos (- \frac{2 + 46}{10}) + 2 πn :

False

arccos (- \frac{2 + 46}{10}) + 2 πn

Plug in

n = 1

arccos (- \frac{2 + 46}{10}) + 2 π 1

For

5 sin (θ) - 5 cos (θ) = 2

plug in

θ = arccos (- \frac{2 + 46}{10}) + 2 π 1

5 sin (arccos (- \frac{2 + 46}{10}) + 2 π 1) - 5 cos (arccos (- \frac{2 + 46}{10}) + 2 π 1) = 2

Refine

6.78232 \dots = 2

\Rightarrow False

Check the solution

- arccos (- \frac{2 + 46}{10}) + 2 πn :

True

- arccos (- \frac{2 + 46}{10}) + 2 πn

Plug in

n = 1

- arccos (- \frac{2 + 46}{10}) + 2 π 1

For

5 sin (θ) - 5 cos (θ) = 2

plug in

θ = - arccos (- \frac{2 + 46}{10}) + 2 π 1

5 sin (- arccos (- \frac{2 + 46}{10}) + 2 π 1) - 5 cos (- arccos (- \frac{2 + 46}{10}) + 2 π 1) = 2

Refine

2 = 2

\Rightarrow True

Check the solution

arccos (\frac{46 - 2}{10}) + 2 πn :

True

arccos (\frac{46 - 2}{10}) + 2 πn

Plug in

n = 1

arccos (\frac{46 - 2}{10}) + 2 π 1

For

5 sin (θ) - 5 cos (θ) = 2

plug in

θ = arccos (\frac{46 - 2}{10}) + 2 π 1

5 sin (arccos (\frac{46 - 2}{10}) + 2 π 1) - 5 cos (arccos (\frac{46 - 2}{10}) + 2 π 1) = 2

Refine

2 = 2

\Rightarrow True

Check the solution

2 π - arccos (\frac{46 - 2}{10}) + 2 πn :

False

2 π - arccos (\frac{46 - 2}{10}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{46 - 2}{10}) + 2 π 1

For

5 sin (θ) - 5 cos (θ) = 2

plug in

θ = 2 π - arccos (\frac{46 - 2}{10}) + 2 π 1

5 sin (2 π - arccos (\frac{46 - 2}{10}) + 2 π 1) - 5 cos (2 π - arccos (\frac{46 - 2}{10}) + 2 π 1) = 2

Refine

- 6.78232 \dots = 2

\Rightarrow False

θ = - arccos (- \frac{2 + 46}{10}) + 2 πn, θ = arccos (\frac{46 - 2}{10}) + 2 πn

Show solutions in decimal form

θ = - 2.64295 \dots + 2 πn, θ = 1.07215 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 5sin(θ)-5cos(θ)=2 ?
The general solution for 5sin(θ)-5cos(θ)=2 is θ=-2.64295…+2pin,θ=1.07215…+2pin