What is the general solution for 4sin(θ)-4cos(θ)=2 ?

The general solution for 4sin(θ)-4cos(θ)=2 is θ=-2.71756…+2pin,θ=1.14676…+2pin

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4sin(θ)-4cos(θ)=2

Solution

4 sin (θ) - 4 cos (θ) = 2

Solution

θ = - 2.71756 \dots + 2 πn, θ = 1.14676 \dots + 2 πn

Degrees

θ = - 155.70481 \dots^{\circ} + 36 0^{\circ} n, θ = 65.70481 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 sin (θ) - 4 cos (θ) = 2

Add

4 cos (θ)

to both sides

4 sin (θ) = 2 + 4 cos (θ)

Square both sides

(4 sin (θ))^{2} = (2 + 4 cos (θ))^{2}

Subtract

(2 + 4 cos (θ))^{2}

from both sides

16 sin^{2} (θ) - 4 - 16 cos (θ) - 16 cos^{2} (θ) = 0

Rewrite using trig identities

- 4 - 16 cos (θ) - 16 cos^{2} (θ) + 16 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= - 4 - 16 cos (θ) - 16 cos^{2} (θ) + 16 (1 - cos^{2} (θ))

Simplify

- 4 - 16 cos (θ) - 16 cos^{2} (θ) + 16 (1 - cos^{2} (θ)) : - 32 cos^{2} (θ) - 16 cos (θ) + 12

- 4 - 16 cos (θ) - 16 cos^{2} (θ) + 16 (1 - cos^{2} (θ))

Expand

16 (1 - cos^{2} (θ)) : 16 - 16 cos^{2} (θ)

16 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 16, b = 1, c = cos^{2} (θ)

= 16 \cdot 1 - 16 cos^{2} (θ)

Multiply the numbers:

16 \cdot 1 = 16

= 16 - 16 cos^{2} (θ)

= - 4 - 16 cos (θ) - 16 cos^{2} (θ) + 16 - 16 cos^{2} (θ)

Simplify

- 4 - 16 cos (θ) - 16 cos^{2} (θ) + 16 - 16 cos^{2} (θ) : - 32 cos^{2} (θ) - 16 cos (θ) + 12

- 4 - 16 cos (θ) - 16 cos^{2} (θ) + 16 - 16 cos^{2} (θ)

Group like terms

= - 16 cos (θ) - 16 cos^{2} (θ) - 16 cos^{2} (θ) - 4 + 16

Add similar elements:

- 16 cos^{2} (θ) - 16 cos^{2} (θ) = - 32 cos^{2} (θ)

= - 16 cos (θ) - 32 cos^{2} (θ) - 4 + 16

Add/Subtract the numbers:

- 4 + 16 = 12

= - 32 cos^{2} (θ) - 16 cos (θ) + 12

= - 32 cos^{2} (θ) - 16 cos (θ) + 12

= - 32 cos^{2} (θ) - 16 cos (θ) + 12

12 - 16 cos (θ) - 32 cos^{2} (θ) = 0

Solve by substitution

12 - 16 cos (θ) - 32 cos^{2} (θ) = 0

Let:

cos (θ) = u

12 - 16 u - 32 u^{2} = 0

12 - 16 u - 32 u^{2} = 0 : u = - \frac{1 + 7}{4}, u = \frac{7 - 1}{4}

12 - 16 u - 32 u^{2} = 0

Write in the standard form

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

- 32 u^{2} - 16 u + 12 = 0

Solve with the quadratic formula

- 32 u^{2} - 16 u + 12 = 0

Quadratic Equation Formula:

For

a = - 32, b = - 16, c = 12

u_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 ( - 32 ) \cdot 12}{2 ( - 32 )}

u_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 ( - 32 ) \cdot 12}{2 ( - 32 )}

(- 16)^{2} - 4 (- 32) \cdot 12 = 167

(- 16)^{2} - 4 (- 32) \cdot 12

Apply rule

- (- a) = a

= (- 16)^{2} + 4 \cdot 32 \cdot 12

Apply exponent rule:

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

n

is even

(- 16)^{2} = 1 6^{2}

= 1 6^{2} + 4 \cdot 32 \cdot 12

Multiply the numbers:

4 \cdot 32 \cdot 12 = 1536

= 1 6^{2} + 1536

1 6^{2} = 256

= 256 + 1536

Add the numbers:

256 + 1536 = 1792

= 1792

Prime factorization of

1792 : 2^{8} \cdot 7

1792

1792

divides by

21792 = 896 \cdot 2

= 2 \cdot 896

896

divides by

2896 = 448 \cdot 2

= 2 \cdot 2 \cdot 448

448

divides by

2448 = 224 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 224

224

divides by

2224 = 112 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 112

112

divides by

2112 = 56 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 56

56

divides by

256 = 28 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 28

28

divides by

228 = 14 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 14

14

divides by

214 = 7 \cdot 2

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

2, 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 7

= 2^{8} \cdot 7

= 2^{8} \cdot 7

Apply radical rule:

= 7 2^{8}

Apply radical rule:

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

= 2^{4} 7

Refine

= 167

u_{1, 2} = \frac{- ( - 16 ) \pm 16 7}{2 ( - 32 )}

Separate the solutions

u_{1} = \frac{- ( - 16 ) + 16 7}{2 ( - 32 )}, u_{2} = \frac{- ( - 16 ) - 16 7}{2 ( - 32 )}

u = \frac{- ( - 16 ) + 16 7}{2 ( - 32 )} : - \frac{1 + 7}{4}

\frac{- ( - 16 ) + 16 7}{2 ( - 32 )}

Remove parentheses:

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

= \frac{16 + 16 7}{- 2 \cdot 32}

Multiply the numbers:

2 \cdot 32 = 64

= \frac{16 + 16 7}{- 64}

Apply the fraction rule:

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

= - \frac{16 + 16 7}{64}

Cancel

\frac{16 + 16 7}{64} : \frac{1 + 7}{4}

\frac{16 + 16 7}{64}

Factor

16 + 167 : 16 (1 + 7)

16 + 167

Rewrite as

= 16 \cdot 1 + 167

Factor out common term

16

= 16 (1 + 7)

= \frac{16 ( 1 + 7 )}{64}

Cancel the common factor:

16

= \frac{1 + 7}{4}

= - \frac{1 + 7}{4}

u = \frac{- ( - 16 ) - 16 7}{2 ( - 32 )} : \frac{7 - 1}{4}

\frac{- ( - 16 ) - 16 7}{2 ( - 32 )}

Remove parentheses:

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

= \frac{16 - 16 7}{- 2 \cdot 32}

Multiply the numbers:

2 \cdot 32 = 64

= \frac{16 - 16 7}{- 64}

Apply the fraction rule:

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

16 - 167 = - (167 - 16)

= \frac{16 7 - 16}{64}

Factor

167 - 16 : 16 (7 - 1)

167 - 16

Rewrite as

= 167 - 16 \cdot 1

Factor out common term

16

= 16 (7 - 1)

= \frac{16 ( 7 - 1 )}{64}

Cancel the common factor:

16

= \frac{7 - 1}{4}

The solutions to the quadratic equation are:

u = - \frac{1 + 7}{4}, u = \frac{7 - 1}{4}

Substitute back

u = cos (θ)

cos (θ) = - \frac{1 + 7}{4}, cos (θ) = \frac{7 - 1}{4}

cos (θ) = - \frac{1 + 7}{4}, cos (θ) = \frac{7 - 1}{4}

cos (θ) = - \frac{1 + 7}{4} : θ = arccos (- \frac{1 + 7}{4}) + 2 πn, θ = - arccos (- \frac{1 + 7}{4}) + 2 πn

cos (θ) = - \frac{1 + 7}{4}

Apply trig inverse properties

cos (θ) = - \frac{1 + 7}{4}

General solutions for

cos (θ) = - \frac{1 + 7}{4}

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

θ = arccos (- \frac{1 + 7}{4}) + 2 πn, θ = - arccos (- \frac{1 + 7}{4}) + 2 πn

θ = arccos (- \frac{1 + 7}{4}) + 2 πn, θ = - arccos (- \frac{1 + 7}{4}) + 2 πn

cos (θ) = \frac{7 - 1}{4} : θ = arccos (\frac{7 - 1}{4}) + 2 πn, θ = 2 π - arccos (\frac{7 - 1}{4}) + 2 πn

cos (θ) = \frac{7 - 1}{4}

Apply trig inverse properties

cos (θ) = \frac{7 - 1}{4}

General solutions for

cos (θ) = \frac{7 - 1}{4}

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

θ = arccos (\frac{7 - 1}{4}) + 2 πn, θ = 2 π - arccos (\frac{7 - 1}{4}) + 2 πn

θ = arccos (\frac{7 - 1}{4}) + 2 πn, θ = 2 π - arccos (\frac{7 - 1}{4}) + 2 πn

Combine all the solutions

θ = arccos (- \frac{1 + 7}{4}) + 2 πn, θ = - arccos (- \frac{1 + 7}{4}) + 2 πn, θ = arccos (\frac{7 - 1}{4}) + 2 πn, θ = 2 π - arccos (\frac{7 - 1}{4}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

4 sin (θ) - 4 cos (θ) = 2

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

Check the solution

arccos (- \frac{1 + 7}{4}) + 2 πn :

False

arccos (- \frac{1 + 7}{4}) + 2 πn

Plug in

n = 1

arccos (- \frac{1 + 7}{4}) + 2 π 1

For

4 sin (θ) - 4 cos (θ) = 2

plug in

θ = arccos (- \frac{1 + 7}{4}) + 2 π 1

4 sin (arccos (- \frac{1 + 7}{4}) + 2 π 1) - 4 cos (arccos (- \frac{1 + 7}{4}) + 2 π 1) = 2

Refine

5.29150 \dots = 2

\Rightarrow False

Check the solution

- arccos (- \frac{1 + 7}{4}) + 2 πn :

True

- arccos (- \frac{1 + 7}{4}) + 2 πn

Plug in

n = 1

- arccos (- \frac{1 + 7}{4}) + 2 π 1

For

4 sin (θ) - 4 cos (θ) = 2

plug in

θ = - arccos (- \frac{1 + 7}{4}) + 2 π 1

4 sin (- arccos (- \frac{1 + 7}{4}) + 2 π 1) - 4 cos (- arccos (- \frac{1 + 7}{4}) + 2 π 1) = 2

Refine

2 = 2

\Rightarrow True

Check the solution

arccos (\frac{7 - 1}{4}) + 2 πn :

True

arccos (\frac{7 - 1}{4}) + 2 πn

Plug in

n = 1

arccos (\frac{7 - 1}{4}) + 2 π 1

For

4 sin (θ) - 4 cos (θ) = 2

plug in

θ = arccos (\frac{7 - 1}{4}) + 2 π 1

4 sin (arccos (\frac{7 - 1}{4}) + 2 π 1) - 4 cos (arccos (\frac{7 - 1}{4}) + 2 π 1) = 2

Refine

2 = 2

\Rightarrow True

Check the solution

2 π - arccos (\frac{7 - 1}{4}) + 2 πn :

False

2 π - arccos (\frac{7 - 1}{4}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{7 - 1}{4}) + 2 π 1

For

4 sin (θ) - 4 cos (θ) = 2

plug in

θ = 2 π - arccos (\frac{7 - 1}{4}) + 2 π 1

4 sin (2 π - arccos (\frac{7 - 1}{4}) + 2 π 1) - 4 cos (2 π - arccos (\frac{7 - 1}{4}) + 2 π 1) = 2

Refine

- 5.29150 \dots = 2

\Rightarrow False

θ = - arccos (- \frac{1 + 7}{4}) + 2 πn, θ = arccos (\frac{7 - 1}{4}) + 2 πn

Show solutions in decimal form

θ = - 2.71756 \dots + 2 πn, θ = 1.14676 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin(θ)-4cos(θ)=2 ?
The general solution for 4sin(θ)-4cos(θ)=2 is θ=-2.71756…+2pin,θ=1.14676…+2pin