What is the general solution for -4+cos(4θ)=(-8+sqrt(3))/2 ?

The general solution for -4+cos(4θ)=(-8+sqrt(3))/2 is θ= pi/(24)+(pin)/2 ,θ= pi/2-pi/(24)+(pin)/2

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-4+cos(4θ)=(-8+sqrt(3))/2

Solution

- 4 + cos (4 θ) = \frac{- 8 + 3}{2}

Solution

θ = \frac{π}{24} + \frac{πn}{2}, θ = \frac{π}{2} - \frac{π}{24} + \frac{πn}{2}

Degrees

θ = 7. 5^{\circ} + 9 0^{\circ} n, θ = 82. 5^{\circ} + 9 0^{\circ} n

Solution steps

- 4 + cos (4 θ) = \frac{- 8 + 3}{2}

Move

4

to the right side

- 4 + cos (4 θ) = \frac{- 8 + 3}{2}

Add

4

to both sides

- 4 + cos (4 θ) + 4 = \frac{- 8 + 3}{2} + 4

Simplify

cos (4 θ) = \frac{- 8 + 3}{2} + 4

cos (4 θ) = \frac{- 8 + 3}{2} + 4

Apply trig inverse properties

cos (4 θ) = \frac{- 8 + 3}{2} + 4

General solutions for

cos (4 θ) = \frac{- 8 + 3}{2} + 4

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

4 θ = arccos (\frac{- 8 + 3}{2} + 4) + 2 πn, 4 θ = 2 π - arccos (\frac{- 8 + 3}{2} + 4) + 2 πn

4 θ = arccos (\frac{- 8 + 3}{2} + 4) + 2 πn, 4 θ = 2 π - arccos (\frac{- 8 + 3}{2} + 4) + 2 πn

Solve

4 θ = arccos (\frac{- 8 + 3}{2} + 4) + 2 πn : θ = \frac{π}{24} + \frac{πn}{2}

4 θ = arccos (\frac{- 8 + 3}{2} + 4) + 2 πn

Simplify

arccos (\frac{- 8 + 3}{2} + 4) + 2 πn : \frac{π}{6} + 2 πn

arccos (\frac{- 8 + 3}{2} + 4) + 2 πn

Use the following trivial identity:

arccos (\frac{3 - 8}{2} + 4) = \frac{π}{6}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{6} + 2 πn

4 θ = \frac{π}{6} + 2 πn

Divide both sides by

4

4 θ = \frac{π}{6} + 2 πn

Divide both sides by

4

\frac{4 θ}{4} = \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 θ}{4} = \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

Divide the numbers:

\frac{4}{4} = 1

= θ

Simplify

\frac{\frac{π}{6}}{4} + \frac{2 πn}{4} : \frac{π}{24} + \frac{πn}{2}

\frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

\frac{\frac{π}{6}}{4} = \frac{π}{24}

\frac{\frac{π}{6}}{4}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 4}

Multiply the numbers:

6 \cdot 4 = 24

= \frac{π}{24}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{π}{24} + \frac{πn}{2}

θ = \frac{π}{24} + \frac{πn}{2}

θ = \frac{π}{24} + \frac{πn}{2}

θ = \frac{π}{24} + \frac{πn}{2}

Solve

4 θ = 2 π - arccos (\frac{- 8 + 3}{2} + 4) + 2 πn : θ = \frac{π}{2} - \frac{π}{24} + \frac{πn}{2}

4 θ = 2 π - arccos (\frac{- 8 + 3}{2} + 4) + 2 πn

Simplify

2 π - arccos (\frac{- 8 + 3}{2} + 4) + 2 πn : 2 π - \frac{π}{6} + 2 πn

2 π - arccos (\frac{- 8 + 3}{2} + 4) + 2 πn

Use the following trivial identity:

arccos (\frac{3 - 8}{2} + 4) = \frac{π}{6}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{π}{6} + 2 πn

4 θ = 2 π - \frac{π}{6} + 2 πn

Divide both sides by

4

4 θ = 2 π - \frac{π}{6} + 2 πn

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

Divide the numbers:

\frac{4}{4} = 1

= θ

Simplify

\frac{2 π}{4} - \frac{\frac{π}{6}}{4} + \frac{2 πn}{4} : \frac{π}{2} - \frac{π}{24} + \frac{πn}{2}

\frac{2 π}{4} - \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

\frac{2 π}{4} = \frac{π}{2}

\frac{2 π}{4}

Cancel the common factor:

2

= \frac{π}{2}

\frac{\frac{π}{6}}{4} = \frac{π}{24}

\frac{\frac{π}{6}}{4}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 4}

Multiply the numbers:

6 \cdot 4 = 24

= \frac{π}{24}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{π}{2} - \frac{π}{24} + \frac{πn}{2}

θ = \frac{π}{2} - \frac{π}{24} + \frac{πn}{2}

θ = \frac{π}{2} - \frac{π}{24} + \frac{πn}{2}

θ = \frac{π}{2} - \frac{π}{24} + \frac{πn}{2}

θ = \frac{π}{24} + \frac{πn}{2}, θ = \frac{π}{2} - \frac{π}{24} + \frac{πn}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for -4+cos(4θ)=(-8+sqrt(3))/2 ?
The general solution for -4+cos(4θ)=(-8+sqrt(3))/2 is θ= pi/(24)+(pin)/2 ,θ= pi/2-pi/(24)+(pin)/2