What is the general solution for 5(sin(θ)+1)=5 ?

The general solution for 5(sin(θ)+1)=5 is θ=2pin,θ=pi+2pin

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5(sin(θ)+1)=5

5 (sin (θ) + 1) = 5

θ = 2 πn, θ = π + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

5 (sin (θ) + 1) = 5

Divide both sides by

5

5 (sin (θ) + 1) = 5

Divide both sides by

5

\frac{5 ( sin ( θ ) + 1 )}{5} = \frac{5}{5}

Simplify

sin (θ) + 1 = 1

sin (θ) + 1 = 1

Move

1

to the right side

sin (θ) + 1 = 1

Subtract

1

from both sides

sin (θ) + 1 - 1 = 1 - 1

Simplify

sin (θ) = 0

sin (θ) = 0

General solutions for

sin (θ) = 0

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 = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

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

2 sin^{2} (θ) - 3 cos (θ) = 0

\frac{1}{2} = sin (2 x)

sin (θ) = 0.4

cos (x) = 3 sin (x)

What is the general solution for 5(sin(θ)+1)=5 ?
The general solution for 5(sin(θ)+1)=5 is θ=2pin,θ=pi+2pin