What is the general solution for 1+cos(θ)=sqrt(3)-sin(θ) ?

The general solution for 1+cos(θ)=sqrt(3)-sin(θ) is θ=0.54408…+2pin-pi/4 ,θ=pi-0.54408…+2pin-pi/4

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1+cos(θ)=sqrt(3)-sin(θ)

Solution

1 + cos (θ) = 3 - sin (θ)

Solution

θ = 0.54408 \dots + 2 πn - \frac{π}{4}, θ = π - 0.54408 \dots + 2 πn - \frac{π}{4}

Degrees

θ = - 13.82604 \dots^{\circ} + 36 0^{\circ} n, θ = 103.82604 \dots^{\circ} + 36 0^{\circ} n

Solution steps

1 + cos (θ) = 3 - sin (θ)

Subtract

3 - sin (θ)

from both sides

1 + cos (θ) - 3 + sin (θ) = 0

Rewrite using trig identities

1 + cos (θ) - 3 + 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})

= 1 - 3 + 2 sin (θ + \frac{π}{4})

1 - 3 + 2 sin (θ + \frac{π}{4}) = 0

Move

1

to the right side

1 - 3 + 2 sin (θ + \frac{π}{4}) = 0

Subtract

1

from both sides

1 - 3 + 2 sin (θ + \frac{π}{4}) - 1 = 0 - 1

Simplify

- 3 + 2 sin (θ + \frac{π}{4}) = - 1

- 3 + 2 sin (θ + \frac{π}{4}) = - 1

Move

3

to the right side

- 3 + 2 sin (θ + \frac{π}{4}) = - 1

Add

3

to both sides

- 3 + 2 sin (θ + \frac{π}{4}) + 3 = - 1 + 3

Simplify

2 sin (θ + \frac{π}{4}) = - 1 + 3

2 sin (θ + \frac{π}{4}) = - 1 + 3

Divide both sides by

2

2 sin (θ + \frac{π}{4}) = - 1 + 3

Divide both sides by

2

\frac{2 sin ( θ + \frac{π}{4} )}{2} = - \frac{1}{2} + \frac{3}{2}

Simplify

\frac{2 sin ( θ + \frac{π}{4} )}{2} = - \frac{1}{2} + \frac{3}{2}

Simplify

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

\frac{2 sin ( θ + \frac{π}{4} )}{2}

Cancel the common factor:

2

= sin (θ + \frac{π}{4})

Simplify

- \frac{1}{2} + \frac{3}{2} : \frac{2 ( - 1 + 3 )}{2}

- \frac{1}{2} + \frac{3}{2}

Apply rule

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

= \frac{- 1 + 3}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{( - 1 + 3 ) 2}{2 2}

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2 ( - 1 + 3 )}{2}

sin (θ + \frac{π}{4}) = \frac{2 ( - 1 + 3 )}{2}

sin (θ + \frac{π}{4}) = \frac{2 ( - 1 + 3 )}{2}

sin (θ + \frac{π}{4}) = \frac{2 ( - 1 + 3 )}{2}

Apply trig inverse properties

sin (θ + \frac{π}{4}) = \frac{2 ( - 1 + 3 )}{2}

General solutions for

sin (θ + \frac{π}{4}) = \frac{2 ( - 1 + 3 )}{2}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ + \frac{π}{4} = arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn, θ + \frac{π}{4} = π - arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn

θ + \frac{π}{4} = arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn, θ + \frac{π}{4} = π - arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn

Solve

θ + \frac{π}{4} = arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn : θ = arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

θ + \frac{π}{4} = arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn

Simplify

arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn : arcsin (\frac{- 1 + 3}{2}) + 2 πn

arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn

\frac{2 ( - 1 + 3 )}{2} = \frac{- 1 + 3}{2}

\frac{2 ( - 1 + 3 )}{2}

Apply radical rule:

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

= \frac{2 ^{\frac{1}{2}} ( 3 - 1 )}{2}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{- 1 + 3}{2 ^{1 - \frac{1}{2}}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{- 1 + 3}{2 ^{\frac{1}{2}}}

Apply radical rule:

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

= \frac{- 1 + 3}{2}

= arcsin (\frac{3 - 1}{2}) + 2 πn

θ + \frac{π}{4} = arcsin (\frac{- 1 + 3}{2}) + 2 πn

Move

\frac{π}{4}

to the right side

θ + \frac{π}{4} = arcsin (\frac{- 1 + 3}{2}) + 2 πn

Subtract

\frac{π}{4}

from both sides

θ + \frac{π}{4} - \frac{π}{4} = arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

Simplify

θ + \frac{π}{4} - \frac{π}{4} = arcsin (\frac{- 1 + 3}{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

arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4} : arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

= arcsin (\frac{2 ( 3 - 1 )}{2}) + 2 πn - \frac{π}{4}

\frac{2 ( - 1 + 3 )}{2} = \frac{- 1 + 3}{2}

\frac{2 ( - 1 + 3 )}{2}

Apply radical rule:

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

= \frac{2 ^{\frac{1}{2}} ( 3 - 1 )}{2}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{- 1 + 3}{2 ^{1 - \frac{1}{2}}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{- 1 + 3}{2 ^{\frac{1}{2}}}

Apply radical rule:

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

= \frac{- 1 + 3}{2}

= arcsin (\frac{3 - 1}{2}) + 2 πn - \frac{π}{4}

Could not simplify further

= arcsin (\frac{3 - 1}{2}) + 2 πn - \frac{π}{4}

θ = arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

θ = arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

θ = arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

Solve

θ + \frac{π}{4} = π - arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn : θ = π - arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

θ + \frac{π}{4} = π - arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn

Simplify

π - arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn : π - arcsin (\frac{- 1 + 3}{2}) + 2 πn

π - arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn

\frac{2 ( - 1 + 3 )}{2} = \frac{- 1 + 3}{2}

\frac{2 ( - 1 + 3 )}{2}

Apply radical rule:

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

= \frac{2 ^{\frac{1}{2}} ( 3 - 1 )}{2}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{- 1 + 3}{2 ^{1 - \frac{1}{2}}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{- 1 + 3}{2 ^{\frac{1}{2}}}

Apply radical rule:

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

= \frac{- 1 + 3}{2}

= π - arcsin (\frac{3 - 1}{2}) + 2 πn

θ + \frac{π}{4} = π - arcsin (\frac{- 1 + 3}{2}) + 2 πn

Move

\frac{π}{4}

to the right side

θ + \frac{π}{4} = π - arcsin (\frac{- 1 + 3}{2}) + 2 πn

Subtract

\frac{π}{4}

from both sides

θ + \frac{π}{4} - \frac{π}{4} = π - arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

Simplify

θ + \frac{π}{4} - \frac{π}{4} = π - arcsin (\frac{- 1 + 3}{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

π - arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4} : π - arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

π - arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

= π - arcsin (\frac{2 ( 3 - 1 )}{2}) + 2 πn - \frac{π}{4}

\frac{2 ( - 1 + 3 )}{2} = \frac{- 1 + 3}{2}

\frac{2 ( - 1 + 3 )}{2}

Apply radical rule:

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

= \frac{2 ^{\frac{1}{2}} ( 3 - 1 )}{2}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{- 1 + 3}{2 ^{1 - \frac{1}{2}}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{- 1 + 3}{2 ^{\frac{1}{2}}}

Apply radical rule:

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

= \frac{- 1 + 3}{2}

= π - arcsin (\frac{3 - 1}{2}) + 2 πn - \frac{π}{4}

Could not simplify further

= π - arcsin (\frac{3 - 1}{2}) + 2 πn - \frac{π}{4}

θ = π - arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

θ = π - arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

θ = π - arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

θ = arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}, θ = π - arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

Show solutions in decimal form

θ = 0.54408 \dots + 2 πn - \frac{π}{4}, θ = π - 0.54408 \dots + 2 πn - \frac{π}{4}

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Frequently Asked Questions (FAQ)

What is the general solution for 1+cos(θ)=sqrt(3)-sin(θ) ?
The general solution for 1+cos(θ)=sqrt(3)-sin(θ) is θ=0.54408…+2pin-pi/4 ,θ=pi-0.54408…+2pin-pi/4