What is the general solution for 4=2+2sin(θ) ?

The general solution for 4=2+2sin(θ) is θ= pi/2+2pin

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4=2+2sin(θ)

4 = 2 + 2 sin (θ)

θ = \frac{π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

4 = 2 + 2 sin (θ)

Switch sides

2 + 2 sin (θ) = 4

Move

2

to the right side

2 + 2 sin (θ) = 4

Subtract

2

from both sides

2 + 2 sin (θ) - 2 = 4 - 2

Simplify

2 sin (θ) = 2

2 sin (θ) = 2

Divide both sides by

2

2 sin (θ) = 2

Divide both sides by

2

\frac{2 sin ( θ )}{2} = \frac{2}{2}

Simplify

sin (θ) = 1

sin (θ) = 1

General solutions for

sin (θ) = 1

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

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

2 sin^{2} (x) - 3 sin (x) - 1 = 0

2 sin (6 x) - 1 = 0

sin^{2} (x) = \frac{( 6 m - 5 )}{5}

sin^{2} (θ) - sin (θ) - 6 = 0

cos (2 θ) - 3 cos (θ) + 2 = 0

What is the general solution for 4=2+2sin(θ) ?
The general solution for 4=2+2sin(θ) is θ= pi/2+2pin