What is the general solution for 2cos(θ)=-2 ?

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

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2cos(θ)=-2

2 cos (θ) = - 2

θ = π + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (θ) = - 2

Divide both sides by

2

2 cos (θ) = - 2

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{- 2}{2}

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

x, cos (2 x) + \frac{2}{cos ( 2 x )} + 3 = 0

sin (t) = \frac{250}{25 0 ^{2} + 30 0 ^{2}}

0 = sin (13 x)

cos (a) = \frac{9}{41}

sin^{2} (θ) - 2 = cos^{2} (θ)

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