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

The general solution for 2+sin(3θ)=3-sin(3θ) is θ= pi/(18)+(2pin)/3 ,θ=(5pi}{18}+\frac{2pin)/3

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2+sin(3θ)=3-sin(3θ)

2 + sin (3 θ) = 3 - sin (3 θ)

θ = \frac{π}{18} + \frac{2 πn}{3}, θ = \frac{5 π}{18} + \frac{2 πn}{3}

Solution steps

2 + sin (3 θ) = 3 - sin (3 θ)

Solve by substitution

sin (3 θ) = \frac{1}{2}

sin (3 θ) = \frac{1}{2} : θ = \frac{π}{18} + \frac{2 πn}{3}, θ = \frac{5 π}{18} + \frac{2 πn}{3}

Combine all the solutions

θ = \frac{π}{18} + \frac{2 πn}{3}, θ = \frac{5 π}{18} + \frac{2 πn}{3}

sin (2 x) = - cos (3 x) + sin (4 x)

sin (x + \frac{π}{2}) = - cos (x)

sin (x) = (\frac{3}{5})

x, sin^{2} (x) = cos^{2} (x)

cos (θ) = 1 + 2 cos (θ)

What is the general solution for 2+sin(3θ)=3-sin(3θ) ?
The general solution for 2+sin(3θ)=3-sin(3θ) is θ= pi/(18)+(2pin)/3 ,θ=(5pi}{18}+\frac{2pin)/3