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

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

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2+2sin(θ)=4

2 + 2 sin (θ) = 4

θ = \frac{π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

- 4 sin (8 x) = 0

- sin (x) + 2 cos (x) = 0

\frac{sin ( θ )}{140} = \frac{sin ( 11 0 ^{\circ} )}{622}

(cosh (x)) = sinh (x)

- 6 tan^{2} (θ) + 4 tan (θ) = - tan (θ) + 1

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