What is the general solution for cos(3A)+sin(A)=0 ?

The general solution for cos(3A)+sin(A)=0 is A=(pi+4pin)/4 ,A=-(pi+4pin)/8

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cos(3A)+sin(A)=0

Solution

cos (3 A) + sin (A) = 0

Solution

A = \frac{π + 4 πn}{4}, A = - \frac{π + 4 πn}{8}

Degrees

A = 4 5^{\circ} + 18 0^{\circ} n, A = - 22. 5^{\circ} - 9 0^{\circ} n

Solution steps

cos (3 A) + sin (A) = 0

Subtract

sin (A)

from both sides

cos (3 A) = - sin (A)

Rewrite using trig identities

cos (3 A) = - sin (A)

Use the following identity:

- sin (x) = sin (- x)

cos (3 A) = sin (- (A))

Use the following identity:

cos (x) = sin (\frac{π}{2} - x)

sin (\frac{π}{2} - 3 A) = sin (- (A))

sin (\frac{π}{2} - 3 A) = sin (- (A))

Apply trig inverse properties

sin (\frac{π}{2} - 3 A) = sin (- (A))

sin (x) = sin (y) \Rightarrow x = y + 2 π n, x = π - y + 2 π n

- (A) = \frac{π}{2} - 3 A + 2 πn, - (A) = π - (\frac{π}{2} - 3 A) + 2 πn

- (A) = \frac{π}{2} - 3 A + 2 πn, - (A) = π - (\frac{π}{2} - 3 A) + 2 πn

- (A) = \frac{π}{2} - 3 A + 2 πn : A = \frac{π + 4 πn}{4}

- (A) = \frac{π}{2} - 3 A + 2 πn

Expand

- (A) : - A

- (A)

Remove parentheses:

(a) = a

= - A

- A = \frac{π}{2} - 3 A + 2 πn

Move

3 A

to the left side

- A = \frac{π}{2} - 3 A + 2 πn

Add

3 A

to both sides

- A + 3 A = \frac{π}{2} - 3 A + 2 πn + 3 A

Simplify

2 A = \frac{π}{2} + 2 πn

2 A = \frac{π}{2} + 2 πn

Divide both sides by

2

2 A = \frac{π}{2} + 2 πn

Divide both sides by

2

\frac{2 A}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 A}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 A}{2} : A

\frac{2 A}{2}

Divide the numbers:

\frac{2}{2} = 1

= A

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π + 4 πn}{4}

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Apply rule

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

= \frac{\frac{π}{2} + 2 πn}{2}

Join

\frac{π}{2} + 2 πn : \frac{π + 4 πn}{2}

\frac{π}{2} + 2 πn

Convert element to fraction:

2 πn = \frac{2 πn 2}{2}

= \frac{π}{2} + \frac{2 πn \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{π + 2 πn \cdot 2}{2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π + 4 πn}{2}

= \frac{\frac{π + 4 πn}{2}}{2}

Apply the fraction rule:

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

= \frac{π + 4 πn}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π + 4 πn}{4}

A = \frac{π + 4 πn}{4}

A = \frac{π + 4 πn}{4}

A = \frac{π + 4 πn}{4}

- (A) = π - (\frac{π}{2} - 3 A) + 2 πn : A = - \frac{π + 4 πn}{8}

- (A) = π - (\frac{π}{2} - 3 A) + 2 πn

Expand

- (A) : - A

- (A)

Remove parentheses:

(a) = a

= - A

Expand

π - (\frac{π}{2} - 3 A) + 2 πn : π - \frac{π}{2} + 3 A + 2 πn

π - (\frac{π}{2} - 3 A) + 2 πn

- (\frac{π}{2} - 3 A) : - \frac{π}{2} + 3 A

- (\frac{π}{2} - 3 A)

Distribute parentheses

= - (\frac{π}{2}) - (- 3 A)

Apply minus-plus rules

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

= - \frac{π}{2} + 3 A

= π - \frac{π}{2} + 3 A + 2 πn

- A = π - \frac{π}{2} + 3 A + 2 πn

Move

3 A

to the left side

- A = π - \frac{π}{2} + 3 A + 2 πn

Subtract

3 A

from both sides

- A - 3 A = π - \frac{π}{2} + 3 A + 2 πn - 3 A

Simplify

- 4 A = π - \frac{π}{2} + 2 πn

- 4 A = π - \frac{π}{2} + 2 πn

Divide both sides by

- 4

- 4 A = π - \frac{π}{2} + 2 πn

Divide both sides by

- 4

\frac{- 4 A}{- 4} = \frac{π}{- 4} - \frac{\frac{π}{2}}{- 4} + \frac{2 πn}{- 4}

Simplify

\frac{- 4 A}{- 4} = \frac{π}{- 4} - \frac{\frac{π}{2}}{- 4} + \frac{2 πn}{- 4}

Simplify

\frac{- 4 A}{- 4} : A

\frac{- 4 A}{- 4}

Apply the fraction rule:

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

= \frac{4 A}{4}

Divide the numbers:

\frac{4}{4} = 1

= A

Simplify

\frac{π}{- 4} - \frac{\frac{π}{2}}{- 4} + \frac{2 πn}{- 4} : - \frac{π + 4 πn}{8}

\frac{π}{- 4} - \frac{\frac{π}{2}}{- 4} + \frac{2 πn}{- 4}

Apply rule

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

= \frac{π - \frac{π}{2} + 2 πn}{- 4}

Apply the fraction rule:

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

= - \frac{π - \frac{π}{2} + 2 πn}{4}

Join

π - \frac{π}{2} + 2 πn : \frac{π + 4 πn}{2}

π - \frac{π}{2} + 2 πn

Convert element to fraction:

π = \frac{π 2}{2}, 2 πn = \frac{2 πn 2}{2}

= \frac{π 2}{2} - \frac{π}{2} + \frac{2 πn \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{π 2 - π + 2 πn \cdot 2}{2}

π 2 - π + 2 πn \cdot 2 = π + 4 πn

π 2 - π + 2 πn \cdot 2

Add similar elements:

2 π - π = π

= π + 2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= π + 4 πn

= \frac{π + 4 πn}{2}

= - \frac{\frac{π + 4 πn}{2}}{4}

Simplify

\frac{\frac{π + 4 πn}{2}}{4} : \frac{π + 4 πn}{8}

\frac{\frac{π + 4 πn}{2}}{4}

Apply the fraction rule:

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

= \frac{π + 4 πn}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{π + 4 πn}{8}

= - \frac{π + 4 πn}{8}

A = - \frac{π + 4 πn}{8}

A = - \frac{π + 4 πn}{8}

A = - \frac{π + 4 πn}{8}

A = \frac{π + 4 πn}{4}, A = - \frac{π + 4 πn}{8}

A = \frac{π + 4 πn}{4}, A = - \frac{π + 4 πn}{8}

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Frequently Asked Questions (FAQ)

What is the general solution for cos(3A)+sin(A)=0 ?
The general solution for cos(3A)+sin(A)=0 is A=(pi+4pin)/4 ,A=-(pi+4pin)/8