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}

Solution steps

sin (θ + 1 5^{\circ}) = \frac{2}{3}

Apply trig inverse properties

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}

Solve

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

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

Show solutions in decimal form

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}

θ, 2 sin (θ) - 1 = 0

- 10 sin (x) = - 52

\frac{sin ( 11 5 ^{\circ} )}{20} = \frac{sin ( b )}{11}

y, \frac{sin ( 7 y )}{7} = \frac{cos ( 2 x )}{2} + \frac{1}{2}

3 cos (\frac{π x}{12}) + 8.4 = 7

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