What is the general solution for sin(θ+pi/4)=0 ?

The general solution for sin(θ+pi/4)=0 is θ=2pin-pi/4 ,θ=pi+2pin-pi/4

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sin(θ+pi/4)=0

sin (θ + \frac{π}{4}) = 0

θ = 2 πn - \frac{π}{4}, θ = π + 2 πn - \frac{π}{4}

Degrees

θ = - 4 5^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 36 0^{\circ} n

Solution steps

sin (θ + \frac{π}{4}) = 0

General solutions for

sin (θ + \frac{π}{4}) = 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 + \frac{π}{4} = 0 + 2 πn, θ + \frac{π}{4} = π + 2 πn

θ + \frac{π}{4} = 0 + 2 πn, θ + \frac{π}{4} = π + 2 πn

Solve

θ + \frac{π}{4} = 0 + 2 πn : θ = 2 πn - \frac{π}{4}

θ + \frac{π}{4} = 0 + 2 πn

0 + 2 πn = 2 πn

θ + \frac{π}{4} = 2 πn

Move

\frac{π}{4}

to the right side

θ + \frac{π}{4} = 2 πn

Subtract

\frac{π}{4}

from both sides

θ + \frac{π}{4} - \frac{π}{4} = 2 πn - \frac{π}{4}

Simplify

θ = 2 πn - \frac{π}{4}

θ = 2 πn - \frac{π}{4}