What is the general solution for -14cos(θ)=-7sqrt(3) ?

The general solution for -14cos(θ)=-7sqrt(3) is θ= pi/6+2pin,θ=(11pi)/6+2pin

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-14cos(θ)=-7sqrt(3)

- 14 cos (θ) = - 73

θ = \frac{π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

- 14 cos (θ) = - 73

Divide both sides by

- 14

- 14 cos (θ) = - 73

Divide both sides by

- 14

\frac{- 14 cos ( θ )}{- 14} = \frac{- 7 3}{- 14}

Simplify

\frac{- 14 cos ( θ )}{- 14} = \frac{- 7 3}{- 14}

Simplify

\frac{- 14 cos ( θ )}{- 14} : cos (θ)

\frac{- 14 cos ( θ )}{- 14}

Apply the fraction rule:

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

= \frac{14 cos ( θ )}{14}

Divide the numbers:

\frac{14}{14} = 1

= cos (θ)

Simplify

\frac{- 7 3}{- 14} : \frac{3}{2}

\frac{- 7 3}{- 14}

Apply the fraction rule:

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

= \frac{7 3}{14}

Cancel the common factor:

7

= \frac{3}{2}

cos (θ) = \frac{3}{2}

cos (θ) = \frac{3}{2}

cos (θ) = \frac{3}{2}

General solutions for

cos (θ) = \frac{3}{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{π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

12 cos^{2} (x) = 9

9 sin (2 x) = 9 cos (x)

sec^{2} (x) + 8 tan (x) = 10

tan (5 x) = 0

4 cos^{2} (θ) - 2 = 0

What is the general solution for -14cos(θ)=-7sqrt(3) ?
The general solution for -14cos(θ)=-7sqrt(3) is θ= pi/6+2pin,θ=(11pi)/6+2pin