What is the general solution for sqrt(2)sin(3x)=1,0<= x<= 4pi ?

The general solution for sqrt(2)sin(3x)=1,0<= x<= 4pi is

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sqrt(2)sin(3x)=1,0<= x<= 4pi

2 sin (3 x) = 1, 0 \leq x \leq 4 π

x = \frac{π}{12}, x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{11 π}{12}, x = \frac{17 π}{12}, x = \frac{19 π}{12}, x = \frac{25 π}{12}, x = \frac{9 π}{4}, x = \frac{11 π}{4}, x = \frac{35 π}{12}

Solution steps

2 sin (3 x) = 1, 0 \leq x \leq 4 π

Divide both sides by

2

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

General solutions for

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

3 x = \frac{π}{4} + 2 πn, 3 x = \frac{3 π}{4} + 2 πn

Solve

3 x = \frac{π}{4} + 2 πn : x = \frac{π}{12} + \frac{2 πn}{3}

Solve

3 x = \frac{3 π}{4} + 2 πn : x = \frac{π}{4} + \frac{2 πn}{3}

x = \frac{π}{12} + \frac{2 πn}{3}, x = \frac{π}{4} + \frac{2 πn}{3}

Solutions for the range

0 \leq x \leq 4 π

x = \frac{π}{12}, x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{11 π}{12}, x = \frac{17 π}{12}, x = \frac{19 π}{12}, x = \frac{25 π}{12}, x = \frac{9 π}{4}, x = \frac{11 π}{4}, x = \frac{35 π}{12}

\frac{4}{cos ( x )} - 6 = tan (x)

tan (2 a) cot (a + 2 0^{\circ}) = 1

2 sin (3 x) - 1 = 0, 0, 2 π

4 cos^{2} (x) = 2 cos (x) + 1

\frac{13}{sin ( 10 8 ^{\circ} )} = \frac{9}{sin ( x )}

What is the general solution for sqrt(2)sin(3x)=1,0<= x<= 4pi ?
The general solution for sqrt(2)sin(3x)=1,0<= x<= 4pi is