What is the general solution for -4cos(θ)+1=5 ?

The general solution for -4cos(θ)+1=5 is θ=pi+2pin

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-4cos(θ)+1=5

- 4 cos (θ) + 1 = 5

θ = π + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

- 4 cos (θ) + 1 = 5

Move

1

to the right side

- 4 cos (θ) + 1 = 5

Subtract

1

from both sides

- 4 cos (θ) + 1 - 1 = 5 - 1

Simplify

- 4 cos (θ) = 4

- 4 cos (θ) = 4

Divide both sides by

- 4

- 4 cos (θ) = 4

Divide both sides by

- 4

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

Simplify

cos (θ) = - 1

cos (θ) = - 1

General solutions for

cos (θ) = - 1

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 = π + 2 πn

θ = π + 2 πn

cos (2 x) = 0.4

arcsin (x) = \frac{5 π}{6}

2 cos^{2} (t) - cos (t) = 1

- 2 sin (t) + cos (2 t) \cdot 2 = 0

tan (a) = \frac{4}{3}

What is the general solution for -4cos(θ)+1=5 ?
The general solution for -4cos(θ)+1=5 is θ=pi+2pin