What is the general solution for sin(θ+15)= 2/3 ?

The general solution for sin(θ+15)= 2/3 is θ=0.72972…+360n-15,θ=180-0.72972…+360n-15

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sin(θ+15)= 2/3

sin (θ + 1 5^{\circ}) = \frac{2}{3}

θ = 0.72972 \dots + 36 0^{\circ} n - 1 5^{\circ}, θ = 18 0^{\circ} - 0.72972 \dots + 36 0^{\circ} n - 1 5^{\circ}

Radians

θ = 0.72972 \dots - \frac{π}{12} + 2 πn, θ = π - 0.72972 \dots - \frac{π}{12} + 2 πn

Solution steps

sin (θ + 1 5^{\circ}) = \frac{2}{3}

Apply trig inverse properties

sin (θ + 1 5^{\circ}) = \frac{2}{3}

General solutions for

sin (θ + 1 5^{\circ}) = \frac{2}{3}

sin (x) = a \Rightarrow x = arcsin (a) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (a) + 36 0^{\circ} n

θ + 1 5^{\circ} = arcsin (\frac{2}{3}) + 36 0^{\circ} n, θ + 1 5^{\circ} = 18 0^{\circ} - arcsin (\frac{2}{3}) + 36 0^{\circ} n

θ + 1 5^{\circ} = arcsin (\frac{2}{3}) + 36 0^{\circ} n, θ + 1 5^{\circ} = 18 0^{\circ} - arcsin (\frac{2}{3}) + 36 0^{\circ} n

Solve

θ + 1 5^{\circ} = arcsin (\frac{2}{3}) + 36 0^{\circ} n : θ = arcsin (\frac{2}{3}) + 36 0^{\circ} n - 1 5^{\circ}

θ + 1 5^{\circ} = arcsin (\frac{2}{3}) + 36 0^{\circ} n

Move

1 5^{\circ}

to the right side

θ + 1 5^{\circ} = arcsin (\frac{2}{3}) + 36 0^{\circ} n

Subtract

1 5^{\circ}

from both sides

θ + 1 5^{\circ} - 1 5^{\circ} = arcsin (\frac{2}{3}) + 36 0^{\circ} n - 1 5^{\circ}

Simplify

θ = arcsin (\frac{2}{3}) + 36 0^{\circ} n - 1 5^{\circ}

θ = arcsin (\frac{2}{3}) + 36 0^{\circ} n - 1 5^{\circ}