What is the general solution for sin(3θ+pi/(18))=1,0<= θ<2pi ?

The general solution for sin(3θ+pi/(18))=1,0<= θ<2pi is θ=(4pi)/(27),θ=(22pi)/(27),θ=(40pi)/(27)

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sin(3θ+pi/(18))=1,0<= θ<2pi

sin (3 θ + \frac{π}{18}) = 1, 0 \leq θ < 2 π

θ = \frac{4 π}{27}, θ = \frac{22 π}{27}, θ = \frac{40 π}{27}

Solution steps

sin (3 θ + \frac{π}{18}) = 1, 0 \leq θ < 2 π

General solutions for

sin (3 θ + \frac{π}{18}) = 1

3 θ + \frac{π}{18} = \frac{π}{2} + 2 πn

Solve

3 θ + \frac{π}{18} = \frac{π}{2} + 2 πn : θ = \frac{2 πn}{3} + \frac{4 π}{27}

θ = \frac{2 πn}{3} + \frac{4 π}{27}

Solutions for the range

0 \leq θ < 2 π

θ = \frac{4 π}{27}, θ = \frac{22 π}{27}, θ = \frac{40 π}{27}

1 = tan (c)

sin (a) = \frac{3}{2}

tan^{2} (x) = 4

sec (x) = 3

2 cos^{2} (x) + 1 = - 3 cos (x)

What is the general solution for sin(3θ+pi/(18))=1,0<= θ<2pi ?
The general solution for sin(3θ+pi/(18))=1,0<= θ<2pi is θ=(4pi)/(27),θ=(22pi)/(27),θ=(40pi)/(27)