What is the general solution for 3sqrt(2)sin(v)+3sqrt(2)cos(v)=3 ?

The general solution for 3sqrt(2)sin(v)+3sqrt(2)cos(v)=3 is v=1.83259…+2pin,v=2pi-0.26179…+2pin

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Popular Trigonometry >

3sqrt(2)sin(v)+3sqrt(2)cos(v)=3

Solution

32 sin (v) + 32 cos (v) = 3

Solution

v = 1.83259 \dots + 2 πn, v = 2 π - 0.26179 \dots + 2 πn

Degrees

v = 10 5^{\circ} + 36 0^{\circ} n, v = 34 5^{\circ} + 36 0^{\circ} n

Solution steps

32 sin (v) + 32 cos (v) = 3

Subtract

32 cos (v)

from both sides

32 sin (v) = 3 - 32 cos (v)

Square both sides

(32 sin (v))^{2} = (3 - 32 cos (v))^{2}

Subtract

(3 - 32 cos (v))^{2}

from both sides

18 sin^{2} (v) - 9 + 182 cos (v) - 18 cos^{2} (v) = 0

Rewrite using trig identities

- 9 - 18 cos^{2} (v) + 18 sin^{2} (v) + 18 cos (v) 2

Use the Pythagorean identity:

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

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

= - 9 - 18 cos^{2} (v) + 18 (1 - cos^{2} (v)) + 18 cos (v) 2

Simplify

- 9 - 18 cos^{2} (v) + 18 (1 - cos^{2} (v)) + 18 cos (v) 2 : 182 cos (v) - 36 cos^{2} (v) + 9

- 9 - 18 cos^{2} (v) + 18 (1 - cos^{2} (v)) + 18 cos (v) 2

= - 9 - 18 cos^{2} (v) + 18 (1 - cos^{2} (v)) + 182 cos (v)

Expand

18 (1 - cos^{2} (v)) : 18 - 18 cos^{2} (v)

18 (1 - cos^{2} (v))

Apply the distributive law:

a (b - c) = ab - a c

a = 18, b = 1, c = cos^{2} (v)

= 18 \cdot 1 - 18 cos^{2} (v)

Multiply the numbers:

18 \cdot 1 = 18

= 18 - 18 cos^{2} (v)

= - 9 - 18 cos^{2} (v) + 18 - 18 cos^{2} (v) + 18 cos (v) 2

Simplify

- 9 - 18 cos^{2} (v) + 18 - 18 cos^{2} (v) + 18 cos (v) 2 : 182 cos (v) - 36 cos^{2} (v) + 9

- 9 - 18 cos^{2} (v) + 18 - 18 cos^{2} (v) + 18 cos (v) 2

Group like terms

= - 18 cos^{2} (v) - 18 cos^{2} (v) + 182 cos (v) - 9 + 18

Add similar elements:

- 18 cos^{2} (v) - 18 cos^{2} (v) = - 36 cos^{2} (v)

= - 36 cos^{2} (v) + 182 cos (v) - 9 + 18

Add/Subtract the numbers:

- 9 + 18 = 9

= 182 cos (v) - 36 cos^{2} (v) + 9

= 182 cos (v) - 36 cos^{2} (v) + 9

= 182 cos (v) - 36 cos^{2} (v) + 9

9 - 36 cos^{2} (v) + 18 cos (v) 2 = 0

Solve by substitution

9 - 36 cos^{2} (v) + 18 cos (v) 2 = 0

Let:

cos (v) = u

9 - 36 u^{2} + 18 u 2 = 0

9 - 36 u^{2} + 18 u 2 = 0 : u = - \frac{- 2 + 6}{4}, u = \frac{2 + 6}{4}

9 - 36 u^{2} + 18 u 2 = 0

Write in the standard form

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

- 36 u^{2} + 182 u + 9 = 0

Solve with the quadratic formula

- 36 u^{2} + 182 u + 9 = 0

Quadratic Equation Formula:

For

a = - 36, b = 182, c = 9

u_{1, 2} = \frac{- 18 2 \pm ( 18 2 ) ^{2} - 4 ( - 36 ) \cdot 9}{2 ( - 36 )}

u_{1, 2} = \frac{- 18 2 \pm ( 18 2 ) ^{2} - 4 ( - 36 ) \cdot 9}{2 ( - 36 )}

(182)^{2} - 4 (- 36) \cdot 9 = 186

(182)^{2} - 4 (- 36) \cdot 9

Apply rule

- (- a) = a

= (182)^{2} + 4 \cdot 36 \cdot 9

(182)^{2} = 1 8^{2} \cdot 2

(182)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 1 8^{2} (2)^{2}

(2)^{2} : 2

Apply radical rule:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 2

= 1 8^{2} \cdot 2

4 \cdot 36 \cdot 9 = 1296

4 \cdot 36 \cdot 9

Multiply the numbers:

4 \cdot 36 \cdot 9 = 1296

= 1296

= 1 8^{2} \cdot 2 + 1296

1 8^{2} \cdot 2 = 648

1 8^{2} \cdot 2

1 8^{2} = 324

= 324 \cdot 2

Multiply the numbers:

324 \cdot 2 = 648

= 648

= 648 + 1296

Add the numbers:

648 + 1296 = 1944

= 1944

Prime factorization of

1944 : 2^{3} \cdot 3^{5}

1944

1944

divides by

21944 = 972 \cdot 2

= 2 \cdot 972

972

divides by

2972 = 486 \cdot 2

= 2 \cdot 2 \cdot 486

486

divides by

2486 = 243 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 243

243

divides by

3243 = 81 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 81

81

divides by

381 = 27 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 27

27

divides by

327 = 9 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 9

9

divides by

39 = 3 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3

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

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

Apply exponent rule:

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

= 3^{4} \cdot 2^{2} \cdot 2 \cdot 3

Apply radical rule:

= 2^{2} 3^{4} 2 \cdot 3

Apply radical rule:

2^{2} = 2

= 2 3^{4} 2 \cdot 3

Apply radical rule:

3^{4} = 3^{\frac{4}{2}} = 3^{2}

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

Refine

= 186

u_{1, 2} = \frac{- 18 2 \pm 18 6}{2 ( - 36 )}

Separate the solutions

u_{1} = \frac{- 18 2 + 18 6}{2 ( - 36 )}, u_{2} = \frac{- 18 2 - 18 6}{2 ( - 36 )}

u = \frac{- 18 2 + 18 6}{2 ( - 36 )} : - \frac{- 2 + 6}{4}

\frac{- 18 2 + 18 6}{2 ( - 36 )}

Remove parentheses:

(- a) = - a

= \frac{- 18 2 + 18 6}{- 2 \cdot 36}

Multiply the numbers:

2 \cdot 36 = 72

= \frac{- 18 2 + 18 6}{- 72}

Apply the fraction rule:

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

= - \frac{- 18 2 + 18 6}{72}

Cancel

\frac{- 18 2 + 18 6}{72} : \frac{6 - 2}{4}

\frac{- 18 2 + 18 6}{72}

Factor out common term

18

= \frac{18 ( - 2 + 6 )}{72}

Cancel the common factor:

18

= \frac{- 2 + 6}{4}

= - \frac{6 - 2}{4}

= - \frac{- 2 + 6}{4}

u = \frac{- 18 2 - 18 6}{2 ( - 36 )} : \frac{2 + 6}{4}

\frac{- 18 2 - 18 6}{2 ( - 36 )}

Remove parentheses:

(- a) = - a

= \frac{- 18 2 - 18 6}{- 2 \cdot 36}

Multiply the numbers:

2 \cdot 36 = 72

= \frac{- 18 2 - 18 6}{- 72}

Apply the fraction rule:

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

- 182 - 186 = - (182 + 186)

= \frac{18 2 + 18 6}{72}

Factor out common term

18

= \frac{18 ( 2 + 6 )}{72}

Cancel the common factor:

18

= \frac{2 + 6}{4}

The solutions to the quadratic equation are:

u = - \frac{- 2 + 6}{4}, u = \frac{2 + 6}{4}

Substitute back

u = cos (v)

cos (v) = - \frac{- 2 + 6}{4}, cos (v) = \frac{2 + 6}{4}

cos (v) = - \frac{- 2 + 6}{4}, cos (v) = \frac{2 + 6}{4}

cos (v) = - \frac{- 2 + 6}{4} : v = arccos (- \frac{- 2 + 6}{4}) + 2 πn, v = - arccos (- \frac{- 2 + 6}{4}) + 2 πn

cos (v) = - \frac{- 2 + 6}{4}

Apply trig inverse properties

cos (v) = - \frac{- 2 + 6}{4}

General solutions for

cos (v) = - \frac{- 2 + 6}{4}

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

v = arccos (- \frac{- 2 + 6}{4}) + 2 πn, v = - arccos (- \frac{- 2 + 6}{4}) + 2 πn

v = arccos (- \frac{- 2 + 6}{4}) + 2 πn, v = - arccos (- \frac{- 2 + 6}{4}) + 2 πn

cos (v) = \frac{2 + 6}{4} : v = arccos (\frac{2 + 6}{4}) + 2 πn, v = 2 π - arccos (\frac{2 + 6}{4}) + 2 πn

cos (v) = \frac{2 + 6}{4}

Apply trig inverse properties

cos (v) = \frac{2 + 6}{4}

General solutions for

cos (v) = \frac{2 + 6}{4}

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

v = arccos (\frac{2 + 6}{4}) + 2 πn, v = 2 π - arccos (\frac{2 + 6}{4}) + 2 πn

v = arccos (\frac{2 + 6}{4}) + 2 πn, v = 2 π - arccos (\frac{2 + 6}{4}) + 2 πn

Combine all the solutions

v = arccos (- \frac{- 2 + 6}{4}) + 2 πn, v = - arccos (- \frac{- 2 + 6}{4}) + 2 πn, v = arccos (\frac{2 + 6}{4}) + 2 πn, v = 2 π - arccos (\frac{2 + 6}{4}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

32 sin (v) + 32 cos (v) = 3

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

Check the solution

arccos (- \frac{- 2 + 6}{4}) + 2 πn :

True

arccos (- \frac{- 2 + 6}{4}) + 2 πn

Plug in

n = 1

arccos (- \frac{- 2 + 6}{4}) + 2 π 1

For

32 sin (v) + 32 cos (v) = 3

plug in

v = arccos (- \frac{- 2 + 6}{4}) + 2 π 1

32 sin (arccos (- \frac{- 2 + 6}{4}) + 2 π 1) + 32 cos (arccos (- \frac{- 2 + 6}{4}) + 2 π 1) = 3

Refine

3 = 3

\Rightarrow True

Check the solution

- arccos (- \frac{- 2 + 6}{4}) + 2 πn :

False

- arccos (- \frac{- 2 + 6}{4}) + 2 πn

Plug in

n = 1

- arccos (- \frac{- 2 + 6}{4}) + 2 π 1

For

32 sin (v) + 32 cos (v) = 3

plug in

v = - arccos (- \frac{- 2 + 6}{4}) + 2 π 1

32 sin (- arccos (- \frac{- 2 + 6}{4}) + 2 π 1) + 32 cos (- arccos (- \frac{- 2 + 6}{4}) + 2 π 1) = 3

Refine

- 5.19615 \dots = 3

\Rightarrow False

Check the solution

arccos (\frac{2 + 6}{4}) + 2 πn :

False

arccos (\frac{2 + 6}{4}) + 2 πn

Plug in

n = 1

arccos (\frac{2 + 6}{4}) + 2 π 1

For

32 sin (v) + 32 cos (v) = 3

plug in

v = arccos (\frac{2 + 6}{4}) + 2 π 1

32 sin (arccos (\frac{2 + 6}{4}) + 2 π 1) + 32 cos (arccos (\frac{2 + 6}{4}) + 2 π 1) = 3

Refine

5.19615 \dots = 3

\Rightarrow False

Check the solution

2 π - arccos (\frac{2 + 6}{4}) + 2 πn :

True

2 π - arccos (\frac{2 + 6}{4}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{2 + 6}{4}) + 2 π 1

For

32 sin (v) + 32 cos (v) = 3

plug in

v = 2 π - arccos (\frac{2 + 6}{4}) + 2 π 1

32 sin (2 π - arccos (\frac{2 + 6}{4}) + 2 π 1) + 32 cos (2 π - arccos (\frac{2 + 6}{4}) + 2 π 1) = 3

Refine

3 = 3

\Rightarrow True

v = arccos (- \frac{- 2 + 6}{4}) + 2 πn, v = 2 π - arccos (\frac{2 + 6}{4}) + 2 πn

Show solutions in decimal form

v = 1.83259 \dots + 2 πn, v = 2 π - 0.26179 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 3sqrt(2)sin(v)+3sqrt(2)cos(v)=3 ?
The general solution for 3sqrt(2)sin(v)+3sqrt(2)cos(v)=3 is v=1.83259…+2pin,v=2pi-0.26179…+2pin