What is the general solution for 2cos(2θ+10)=1 ?

The general solution for 2cos(2θ+10)=1 is θ=180n+25,θ=180n+145

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Popular Trigonometry >

2cos(2θ+10)=1

Solution

2 cos (2 θ + 1 0^{\circ}) = 1

Solution

θ = 18 0^{\circ} n + 2 5^{\circ}, θ = 18 0^{\circ} n + 14 5^{\circ}

Radians

θ = \frac{5 π}{36} + πn, θ = \frac{29 π}{36} + πn

Solution steps

2 cos (2 θ + 1 0^{\circ}) = 1

Divide both sides by

2

2 cos (2 θ + 1 0^{\circ}) = 1

Divide both sides by

2

\frac{2 cos ( 2 θ + 1 0 ^{\circ} )}{2} = \frac{1}{2}

Simplify

cos (2 θ + 1 0^{\circ}) = \frac{1}{2}

cos (2 θ + 1 0^{\circ}) = \frac{1}{2}

General solutions for

cos (2 θ + 1 0^{\circ}) = \frac{1}{2}

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 θ + 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n, 2 θ + 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n

2 θ + 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n, 2 θ + 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n

Solve

2 θ + 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n : θ = 18 0^{\circ} n + 2 5^{\circ}

2 θ + 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n

Move

1 0^{\circ}

to the right side

2 θ + 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n

Subtract

1 0^{\circ}

from both sides

2 θ + 1 0^{\circ} - 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

Simplify

2 θ + 1 0^{\circ} - 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

Simplify

2 θ + 1 0^{\circ} - 1 0^{\circ} : 2 θ

2 θ + 1 0^{\circ} - 1 0^{\circ}

Add similar elements:

1 0^{\circ} - 1 0^{\circ} = 0

= 2 θ

Simplify

6 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ} : 36 0^{\circ} n + 5 0^{\circ}

6 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

Group like terms

= 36 0^{\circ} n + 6 0^{\circ} - 1 0^{\circ}

Least Common Multiplier of

3, 18 : 18

3, 18

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

18 : 2 \cdot 3 \cdot 3

18

18

divides by

218 = 9 \cdot 2

= 2 \cdot 9

9

divides by

39 = 3 \cdot 3

= 2 \cdot 3 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3 \cdot 3

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

3

18

= 3 \cdot 3 \cdot 2

Multiply the numbers:

3 \cdot 3 \cdot 2 = 18

= 18

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

18

For

6 0^{\circ} :

multiply the denominator and numerator by

6

6 0^{\circ} = \frac{18 0 ^{\circ} 6}{3 \cdot 6} = 6 0^{\circ}

= 6 0^{\circ} - 1 0^{\circ}

Since the denominators are equal, combine the fractions:

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

= \frac{18 0 ^{\circ} 6 - 18 0 ^{\circ}}{18}

Add similar elements:

108 0^{\circ} - 18 0^{\circ} = 90 0^{\circ}

= 36 0^{\circ} n + 5 0^{\circ}

2 θ = 36 0^{\circ} n + 5 0^{\circ}

2 θ = 36 0^{\circ} n + 5 0^{\circ}

2 θ = 36 0^{\circ} n + 5 0^{\circ}

Divide both sides by

2

2 θ = 36 0^{\circ} n + 5 0^{\circ}

Divide both sides by

2

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2} + \frac{5 0 ^{\circ}}{2}

Simplify

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2} + \frac{5 0 ^{\circ}}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{36 0 ^{\circ} n}{2} + \frac{5 0 ^{\circ}}{2} : 18 0^{\circ} n + 2 5^{\circ}

\frac{36 0 ^{\circ} n}{2} + \frac{5 0 ^{\circ}}{2}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

\frac{5 0 ^{\circ}}{2} = 2 5^{\circ}

\frac{5 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{90 0 ^{\circ}}{18 \cdot 2}

Multiply the numbers:

18 \cdot 2 = 36

= 2 5^{\circ}

= 18 0^{\circ} n + 2 5^{\circ}

θ = 18 0^{\circ} n + 2 5^{\circ}

θ = 18 0^{\circ} n + 2 5^{\circ}

θ = 18 0^{\circ} n + 2 5^{\circ}

Solve

2 θ + 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n : θ = 18 0^{\circ} n + 14 5^{\circ}

2 θ + 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n

Move

1 0^{\circ}

to the right side

2 θ + 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n

Subtract

1 0^{\circ}

from both sides

2 θ + 1 0^{\circ} - 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

Simplify

2 θ + 1 0^{\circ} - 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

Simplify

2 θ + 1 0^{\circ} - 1 0^{\circ} : 2 θ

2 θ + 1 0^{\circ} - 1 0^{\circ}

Add similar elements:

1 0^{\circ} - 1 0^{\circ} = 0

= 2 θ

Simplify

30 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ} : 36 0^{\circ} n + 29 0^{\circ}

30 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

Group like terms

= 36 0^{\circ} n + 30 0^{\circ} - 1 0^{\circ}

Least Common Multiplier of

3, 18 : 18

3, 18

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

18 : 2 \cdot 3 \cdot 3

18

18

divides by

218 = 9 \cdot 2

= 2 \cdot 9

9

divides by

39 = 3 \cdot 3

= 2 \cdot 3 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3 \cdot 3

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

3

18

= 3 \cdot 3 \cdot 2

Multiply the numbers:

3 \cdot 3 \cdot 2 = 18

= 18

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

18

For

30 0^{\circ} :

multiply the denominator and numerator by

6

30 0^{\circ} = \frac{90 0 ^{\circ} 6}{3 \cdot 6} = 30 0^{\circ}

= 30 0^{\circ} - 1 0^{\circ}

Since the denominators are equal, combine the fractions:

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

= \frac{540 0 ^{\circ} - 18 0 ^{\circ}}{18}

Add similar elements:

540 0^{\circ} - 18 0^{\circ} = 522 0^{\circ}

= 36 0^{\circ} n + 29 0^{\circ}

2 θ = 36 0^{\circ} n + 29 0^{\circ}

2 θ = 36 0^{\circ} n + 29 0^{\circ}

2 θ = 36 0^{\circ} n + 29 0^{\circ}

Divide both sides by

2

2 θ = 36 0^{\circ} n + 29 0^{\circ}

Divide both sides by

2

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2} + \frac{29 0 ^{\circ}}{2}

Simplify

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2} + \frac{29 0 ^{\circ}}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{36 0 ^{\circ} n}{2} + \frac{29 0 ^{\circ}}{2} : 18 0^{\circ} n + 14 5^{\circ}

\frac{36 0 ^{\circ} n}{2} + \frac{29 0 ^{\circ}}{2}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

\frac{29 0 ^{\circ}}{2} = 14 5^{\circ}

\frac{29 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{522 0 ^{\circ}}{18 \cdot 2}

Multiply the numbers:

18 \cdot 2 = 36

= 14 5^{\circ}

= 18 0^{\circ} n + 14 5^{\circ}

θ = 18 0^{\circ} n + 14 5^{\circ}

θ = 18 0^{\circ} n + 14 5^{\circ}

θ = 18 0^{\circ} n + 14 5^{\circ}

θ = 18 0^{\circ} n + 2 5^{\circ}, θ = 18 0^{\circ} n + 14 5^{\circ}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2θ+10)=1 ?
The general solution for 2cos(2θ+10)=1 is θ=180n+25,θ=180n+145