What is the general solution for -4cos(θ)+1=5,0<= θ<2pi ?

The general solution for -4cos(θ)+1=5,0<= θ<2pi is θ=pi

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-4cos(θ)+1=5,0<= θ<2pi

- 4 cos (θ) + 1 = 5, 0 \leq θ < 2 π

θ = π

Degrees

θ = 18 0^{\circ}

Solution steps

- 4 cos (θ) + 1 = 5, 0 \leq θ < 2 π

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

Solutions for the range

0 \leq θ < 2 π

θ = π

cos (2 θ) = 12 (1 + cos (2 θ))

0 = e^{t a n (x)} - 2

cot (x) = \frac{3}{2}

4 tan^{2} (x) - 1 = 0

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

What is the general solution for -4cos(θ)+1=5,0<= θ<2pi ?
The general solution for -4cos(θ)+1=5,0<= θ<2pi is θ=pi