What is the general solution for -6cos(θ)-10=-7 ?

The general solution for -6cos(θ)-10=-7 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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-6cos(θ)-10=-7

- 6 cos (θ) - 10 = - 7

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

Degrees

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

- 6 cos (θ) - 10 = - 7

Move

10

to the right side

- 6 cos (θ) - 10 = - 7

Add

10

to both sides

- 6 cos (θ) - 10 + 10 = - 7 + 10

Simplify

- 6 cos (θ) = 3

- 6 cos (θ) = 3

Divide both sides by

- 6

- 6 cos (θ) = 3

Divide both sides by

- 6

\frac{- 6 cos ( θ )}{- 6} = \frac{3}{- 6}

Simplify

\frac{- 6 cos ( θ )}{- 6} = \frac{3}{- 6}

Simplify

\frac{- 6 cos ( θ )}{- 6} : cos (θ)

\frac{- 6 cos ( θ )}{- 6}

Apply the fraction rule:

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

= \frac{6 cos ( θ )}{6}

Divide the numbers:

\frac{6}{6} = 1

= cos (θ)

Simplify

\frac{3}{- 6} : - \frac{1}{2}

\frac{3}{- 6}

Apply the fraction rule:

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

= - \frac{3}{6}

Cancel the common factor:

3

= - \frac{1}{2}

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2}

General solutions for

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

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

sin^{2} (x) = cos^{2} (x) + 1

tanh (x) = 1

tan (θ) = \frac{1}{3}

sin (θ) = \frac{20}{29}

2 cos (\frac{1}{2} x) - 3 = 0

What is the general solution for -6cos(θ)-10=-7 ?
The general solution for -6cos(θ)-10=-7 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin