What is the general solution for 4cos(θ)+2sqrt(2)=0 ?

The general solution for 4cos(θ)+2sqrt(2)=0 is θ=(3pi)/4+2pin,θ=(5pi)/4+2pin

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4cos(θ)+2sqrt(2)=0

4 cos (θ) + 22 = 0

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

4 cos (θ) + 22 = 0

Move

22

to the right side

4 cos (θ) + 22 = 0

Subtract

22

from both sides

4 cos (θ) + 22 - 22 = 0 - 22

Simplify

4 cos (θ) = - 22

4 cos (θ) = - 22

Divide both sides by

4

4 cos (θ) = - 22

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 cos ( θ )}{4} : cos (θ)

\frac{4 cos ( θ )}{4}

Divide the numbers:

\frac{4}{4} = 1

= cos (θ)

Simplify

\frac{- 2 2}{4} : - \frac{2}{2}

\frac{- 2 2}{4}

Apply the fraction rule:

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

= - \frac{2 2}{4}

Cancel the common factor:

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

tan^{2} (x) + 1 = 0

sec (θ) = \frac{- 2 3}{3}

2 sin^{2} (x) - 3 sin (x) + 1 = 0, 0 \leq x \leq 2 π

16 sin^{2} (x) - 1 = 0

4 cos^{2} (x) + 4 cos (x) = - 1

What is the general solution for 4cos(θ)+2sqrt(2)=0 ?
The general solution for 4cos(θ)+2sqrt(2)=0 is θ=(3pi)/4+2pin,θ=(5pi)/4+2pin