What is the general solution for -2=2sqrt(2)cos(θ) ?

The general solution for -2=2sqrt(2)cos(θ) is θ=(3pi)/4+2pin,θ=(5pi)/4+2pin

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-2=2sqrt(2)cos(θ)

- 2 = 22 cos (θ)

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

Degrees

θ = 13 5^{\circ} + 36 0^{\circ} n, θ = 22 5^{\circ} + 36 0^{\circ} n

Solution steps

- 2 = 22 cos (θ)

Switch sides

22 cos (θ) = - 2

Divide both sides by

22

22 cos (θ) = - 2

Divide both sides by

22

\frac{2 2 cos ( θ )}{2 2} = \frac{- 2}{2 2}

Simplify

\frac{2 2 cos ( θ )}{2 2} = \frac{- 2}{2 2}

Simplify

\frac{2 2 cos ( θ )}{2 2} : cos (θ)

\frac{2 2 cos ( θ )}{2 2}

Divide the numbers:

\frac{2}{2} = 1

= \frac{2 cos ( θ )}{2}

Cancel the common factor:

2

= cos (θ)

Simplify

\frac{- 2}{2 2} : - \frac{2}{2}

\frac{- 2}{2 2}

Apply the fraction rule:

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

= - \frac{2}{2 2}

Divide the numbers:

\frac{2}{2} = 1

= - \frac{1}{2}

Rationalize

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

General solutions for

cos (θ) = - \frac{2}{2}

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}

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

cos^{2} (2 x) = 1

- 4 cos^{2} (θ) = - 1

- sin (2 x) + sin (x) = 0

sin^{2} (x) = 3 (1 - cos (x))

cos (x) = \frac{1}{10}

What is the general solution for -2=2sqrt(2)cos(θ) ?
The general solution for -2=2sqrt(2)cos(θ) is θ=(3pi)/4+2pin,θ=(5pi)/4+2pin