What is the general solution for ((sqrt(2))/2)(cos(θ)+sin(θ))=1 ?

The general solution for ((sqrt(2))/2)(cos(θ)+sin(θ))=1 is θ=2pin+pi/4

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((sqrt(2))/2)(cos(θ)+sin(θ))=1

Solution

(\frac{2}{2}) (cos (θ) + sin (θ)) = 1

Solution

θ = 2 πn + \frac{π}{4}

Degrees

θ = 4 5^{\circ} + 36 0^{\circ} n

Solution steps

(\frac{2}{2}) (cos (θ) + sin (θ)) = 1

Rewrite using trig identities

\frac{2}{2} (cos (θ) + sin (θ))

sin (θ) + cos (θ) = 2 sin (θ + \frac{π}{4})

sin (θ) + cos (θ)

Rewrite as

= 2 (\frac{1}{2} sin (θ) + \frac{1}{2} cos (θ))

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (θ) + sin (\frac{π}{4}) cos (θ))

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (θ + \frac{π}{4})

= \frac{2}{2} 2 sin (θ + \frac{π}{4})

\frac{2}{2} 2 sin (θ + \frac{π}{4}) = 1

Simplify

\frac{2}{2} 2 sin (θ + \frac{π}{4}) : sin (θ + \frac{π}{4})

\frac{2}{2} 2 sin (θ + \frac{π}{4})

Multiply fractions:

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

= \frac{2 2 sin ( θ + \frac{π}{4} )}{2}

22 sin (θ + \frac{π}{4}) = 2 sin (θ + \frac{π}{4})

22 sin (θ + \frac{π}{4})

Apply radical rule:

a a = a

22 = 2

= 2 sin (θ + \frac{π}{4})

= \frac{2 sin ( θ + \frac{π}{4} )}{2}

Divide the numbers:

\frac{2}{2} = 1

= sin (θ + \frac{π}{4})

sin (θ + \frac{π}{4}) = 1

General solutions for

sin (θ + \frac{π}{4}) = 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ + \frac{π}{4} = \frac{π}{2} + 2 πn

θ + \frac{π}{4} = \frac{π}{2} + 2 πn

Solve

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

θ + \frac{π}{4} = \frac{π}{2} + 2 πn

Move

\frac{π}{4}

to the right side

θ + \frac{π}{4} = \frac{π}{2} + 2 πn

Subtract

\frac{π}{4}

from both sides

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

Simplify

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

Simplify

θ + \frac{π}{4} - \frac{π}{4} : θ

θ + \frac{π}{4} - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} = 0

= θ

Simplify

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

\frac{π}{2} + 2 πn - \frac{π}{4}

Group like terms

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

Add similar elements:

2 π - π = π

= 2 πn + \frac{π}{4}

θ = 2 πn + \frac{π}{4}

θ = 2 πn + \frac{π}{4}

θ = 2 πn + \frac{π}{4}

θ = 2 πn + \frac{π}{4}

Popular Examples

-6cos^2(θ)+14cos(θ)+15=3cos(θ)+8

- 6 cos^{2} (θ) + 14 cos (θ) + 15 = 3 cos (θ) + 8

cos(θ)=(-1)/(sqrt(2)),0<= θ<= 4pi

cos (θ) = \frac{- 1}{2}, 0 \leq θ \leq 4 π

2sin(θ)=3

2 sin (θ) = 3

sqrt(2)cos(θ)=1,0<= θ<= 360

2 cos (θ) = 1, 0^{\circ} \leq θ \leq 36 0^{\circ}

-2cos(x)=1

- 2 cos (x) = 1

Frequently Asked Questions (FAQ)

What is the general solution for ((sqrt(2))/2)(cos(θ)+sin(θ))=1 ?
The general solution for ((sqrt(2))/2)(cos(θ)+sin(θ))=1 is θ=2pin+pi/4