What is the general solution for 2sin(3x+1)+1=1 ?

The general solution for 2sin(3x+1)+1=1 is x=(2pin)/3-1/3 ,x= pi/3+(2pin)/3-1/3

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2sin(3x+1)+1=1

2 sin (3 x + 1) + 1 = 1

x = \frac{2 πn}{3} - \frac{1}{3}, x = \frac{π}{3} + \frac{2 πn}{3} - \frac{1}{3}

Solution steps

2 sin (3 x + 1) + 1 = 1

Subtract

1

from both sides

2 sin (3 x + 1) + 1 - 1 = 1 - 1

Simplify

2 sin (3 x + 1) = 0

Divide both sides by

2

sin (3 x + 1) = 0

General solutions for

sin (3 x + 1) = 0

3 x + 1 = 0 + 2 πn, 3 x + 1 = π + 2 πn

Solve

3 x + 1 = 0 + 2 πn : x = \frac{2 πn}{3} - \frac{1}{3}

Solve

3 x + 1 = π + 2 πn : x = \frac{π}{3} + \frac{2 πn}{3} - \frac{1}{3}

x = \frac{2 πn}{3} - \frac{1}{3}, x = \frac{π}{3} + \frac{2 πn}{3} - \frac{1}{3}

sin (x) = 6946

sin (2 x) - \frac{3}{2} = 0, - 2 π \leq x \leq 2 π

\frac{sec ( x ) + tan ( x )}{cot ( x ) + cos ( x )} = sec^{2} (x)

cos (x) = csc (x)

1 + cos (2 t) = cos^{2} (t)

What is the general solution for 2sin(3x+1)+1=1 ?
The general solution for 2sin(3x+1)+1=1 is x=(2pin)/3-1/3 ,x= pi/3+(2pin)/3-1/3