What is the general solution for 3+1/5 cos(θ/3+(7pi)/4)=3 ?

The general solution for 3+1/5 cos(θ/3+(7pi)/4)=3 is θ=6pin-(15pi)/4 ,θ=6pin-(3pi)/4

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3+1/5 cos(θ/3+(7pi)/4)=3

Solution

3 + \frac{1}{5} cos (\frac{θ}{3} + \frac{7 π}{4}) = 3

Solution

θ = 6 πn - \frac{15 π}{4}, θ = 6 πn - \frac{3 π}{4}

Degrees

θ = - 67 5^{\circ} + 108 0^{\circ} n, θ = - 13 5^{\circ} + 108 0^{\circ} n

Solution steps

3 + \frac{1}{5} cos (\frac{θ}{3} + \frac{7 π}{4}) = 3

Subtract

3

from both sides

3 + \frac{1}{5} cos (\frac{θ}{3} + \frac{7 π}{4}) - 3 = 3 - 3

Simplify

\frac{1}{5} cos (\frac{θ}{3} + \frac{7 π}{4}) = 0

Multiply both sides by

5

\frac{1}{5} cos (\frac{θ}{3} + \frac{7 π}{4}) = 0

Multiply both sides by

5

5 \cdot \frac{1}{5} cos (\frac{θ}{3} + \frac{7 π}{4}) = 0 \cdot 5

Simplify

cos (\frac{θ}{3} + \frac{7 π}{4}) = 0

cos (\frac{θ}{3} + \frac{7 π}{4}) = 0

General solutions for

cos (\frac{θ}{3} + \frac{7 π}{4}) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

\frac{θ}{3} + \frac{7 π}{4} = \frac{π}{2} + 2 πn, \frac{θ}{3} + \frac{7 π}{4} = \frac{3 π}{2} + 2 πn

\frac{θ}{3} + \frac{7 π}{4} = \frac{π}{2} + 2 πn, \frac{θ}{3} + \frac{7 π}{4} = \frac{3 π}{2} + 2 πn

Solve

\frac{θ}{3} + \frac{7 π}{4} = \frac{π}{2} + 2 πn : θ = 6 πn - \frac{15 π}{4}

\frac{θ}{3} + \frac{7 π}{4} = \frac{π}{2} + 2 πn

Move

\frac{7 π}{4}

to the right side

\frac{θ}{3} + \frac{7 π}{4} = \frac{π}{2} + 2 πn

Subtract

\frac{7 π}{4}

from both sides

\frac{θ}{3} + \frac{7 π}{4} - \frac{7 π}{4} = \frac{π}{2} + 2 πn - \frac{7 π}{4}

Simplify

\frac{θ}{3} + \frac{7 π}{4} - \frac{7 π}{4} = \frac{π}{2} + 2 πn - \frac{7 π}{4}

Simplify

\frac{θ}{3} + \frac{7 π}{4} - \frac{7 π}{4} : \frac{θ}{3}

\frac{θ}{3} + \frac{7 π}{4} - \frac{7 π}{4}

Add similar elements:

\frac{7 π}{4} - \frac{7 π}{4} = 0

= \frac{θ}{3}

Simplify

\frac{π}{2} + 2 πn - \frac{7 π}{4} : 2 πn - \frac{5 π}{4}

\frac{π}{2} + 2 πn - \frac{7 π}{4}

Group like terms

= 2 πn + \frac{π}{2} - \frac{7 π}{4}

Least Common Multiplier of

2, 4 : 4

2, 4

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

2

4

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

4

For

\frac{π}{2} :

multiply the denominator and numerator by

2

\frac{π}{2} = \frac{π 2}{2 \cdot 2} = \frac{π 2}{4}

= \frac{π 2}{4} - \frac{7 π}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{π 2 - 7 π}{4}

Add similar elements:

2 π - 7 π = - 5 π

= \frac{- 5 π}{4}

Apply the fraction rule:

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

= 2 πn - \frac{5 π}{4}

\frac{θ}{3} = 2 πn - \frac{5 π}{4}

\frac{θ}{3} = 2 πn - \frac{5 π}{4}

\frac{θ}{3} = 2 πn - \frac{5 π}{4}

Multiply both sides by

3

\frac{θ}{3} = 2 πn - \frac{5 π}{4}

Multiply both sides by

3

\frac{3 θ}{3} = 3 \cdot 2 πn - 3 \cdot \frac{5 π}{4}

Simplify

\frac{3 θ}{3} = 3 \cdot 2 πn - 3 \cdot \frac{5 π}{4}

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

3 \cdot 2 πn - 3 \cdot \frac{5 π}{4} : 6 πn - \frac{15 π}{4}

3 \cdot 2 πn - 3 \cdot \frac{5 π}{4}

3 \cdot 2 πn = 6 πn

3 \cdot 2 πn

Multiply the numbers:

3 \cdot 2 = 6

= 6 πn

3 \cdot \frac{5 π}{4} = \frac{15 π}{4}

3 \cdot \frac{5 π}{4}

Multiply fractions:

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

= \frac{5 π 3}{4}

Multiply the numbers:

5 \cdot 3 = 15

= \frac{15 π}{4}

= 6 πn - \frac{15 π}{4}

θ = 6 πn - \frac{15 π}{4}

θ = 6 πn - \frac{15 π}{4}

θ = 6 πn - \frac{15 π}{4}

Solve

\frac{θ}{3} + \frac{7 π}{4} = \frac{3 π}{2} + 2 πn : θ = 6 πn - \frac{3 π}{4}

\frac{θ}{3} + \frac{7 π}{4} = \frac{3 π}{2} + 2 πn

Move

\frac{7 π}{4}

to the right side

\frac{θ}{3} + \frac{7 π}{4} = \frac{3 π}{2} + 2 πn

Subtract

\frac{7 π}{4}

from both sides

\frac{θ}{3} + \frac{7 π}{4} - \frac{7 π}{4} = \frac{3 π}{2} + 2 πn - \frac{7 π}{4}

Simplify

\frac{θ}{3} + \frac{7 π}{4} - \frac{7 π}{4} = \frac{3 π}{2} + 2 πn - \frac{7 π}{4}

Simplify

\frac{θ}{3} + \frac{7 π}{4} - \frac{7 π}{4} : \frac{θ}{3}

\frac{θ}{3} + \frac{7 π}{4} - \frac{7 π}{4}

Add similar elements:

\frac{7 π}{4} - \frac{7 π}{4} = 0

= \frac{θ}{3}

Simplify

\frac{3 π}{2} + 2 πn - \frac{7 π}{4} : 2 πn - \frac{π}{4}

\frac{3 π}{2} + 2 πn - \frac{7 π}{4}

Group like terms

= 2 πn + \frac{3 π}{2} - \frac{7 π}{4}

Least Common Multiplier of

2, 4 : 4

2, 4

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

2

4

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

4

For

\frac{3 π}{2} :

multiply the denominator and numerator by

2

\frac{3 π}{2} = \frac{3 π 2}{2 \cdot 2} = \frac{6 π}{4}

= \frac{6 π}{4} - \frac{7 π}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{6 π - 7 π}{4}

Add similar elements:

6 π - 7 π = - π

= \frac{- π}{4}

Apply the fraction rule:

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

= 2 πn - \frac{π}{4}

\frac{θ}{3} = 2 πn - \frac{π}{4}

\frac{θ}{3} = 2 πn - \frac{π}{4}

\frac{θ}{3} = 2 πn - \frac{π}{4}

Multiply both sides by

3

\frac{θ}{3} = 2 πn - \frac{π}{4}

Multiply both sides by

3

\frac{3 θ}{3} = 3 \cdot 2 πn - 3 \cdot \frac{π}{4}

Simplify

\frac{3 θ}{3} = 3 \cdot 2 πn - 3 \cdot \frac{π}{4}

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

3 \cdot 2 πn - 3 \cdot \frac{π}{4} : 6 πn - \frac{3 π}{4}

3 \cdot 2 πn - 3 \cdot \frac{π}{4}

Multiply the numbers:

3 \cdot 2 = 6

= 6 πn - 3 \cdot \frac{π}{4}

Multiply

3 \cdot \frac{π}{4} : \frac{3 π}{4}

3 \cdot \frac{π}{4}

Multiply fractions:

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

= \frac{π 3}{4}

= 6 πn - \frac{3 π}{4}

θ = 6 πn - \frac{3 π}{4}

θ = 6 πn - \frac{3 π}{4}

θ = 6 πn - \frac{3 π}{4}

θ = 6 πn - \frac{15 π}{4}, θ = 6 πn - \frac{3 π}{4}

Popular Examples

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sin (θ) = - \frac{13}{7}

-2sin(x)=-sqrt(2)

- 2 sin (x) = - 2

sin(pi-x)=cos((3pi)/2-x)+cos(pi)

sin (π - x) = cos (\frac{3 π}{2} - x) + cos (π)

4csc(x)+8=0

4 csc (x) + 8 = 0

2cos(2θ)+3=2

2 cos (2 θ) + 3 = 2

Frequently Asked Questions (FAQ)

What is the general solution for 3+1/5 cos(θ/3+(7pi)/4)=3 ?
The general solution for 3+1/5 cos(θ/3+(7pi)/4)=3 is θ=6pin-(15pi)/4 ,θ=6pin-(3pi)/4