What is the general solution for 2sin(6x)sin(3x)=1-cos(9x) ?

The general solution for 2sin(6x)sin(3x)=1-cos(9x) is x=(2pin)/3

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2sin(6x)sin(3x)=1-cos(9x)

2 sin (6 x) sin (3 x) = 1 - cos (9 x)

x = \frac{2 πn}{3}

Solution steps

2 sin (6 x) sin (3 x) = 1 - cos (9 x)

Subtract

1 - cos (9 x)

from both sides

2 sin (6 x) sin (3 x) - 1 + cos (9 x) = 0

Rewrite using trig identities

- 1 + cos (3 x) = 0

Move

1

to the right side

cos (3 x) = 1

General solutions for

cos (3 x) = 1

3 x = 0 + 2 πn

Solve

3 x = 0 + 2 πn : x = \frac{2 πn}{3}

x = \frac{2 πn}{3}

cos (θ) sin (θ) = 172245

sec (b) cot (b) = sin (b)

\frac{37}{sin ( x )} = \frac{30}{sin ( 3 0 ^{\circ} )}

\frac{1}{cos ( θ )} = 1

\frac{sin ( x ) + cos ( x )}{sin ( x )} = \frac{1 + 1}{tan ( x )}

What is the general solution for 2sin(6x)sin(3x)=1-cos(9x) ?
The general solution for 2sin(6x)sin(3x)=1-cos(9x) is x=(2pin)/3