What is the general solution for 2sin(x)cos(x)-sin(2x)cos(2x)=0 ?

The general solution for 2sin(x)cos(x)-sin(2x)cos(2x)=0 is x=pin,x= pi/2+pin

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2sin(x)cos(x)-sin(2x)cos(2x)=0

2 sin (x) cos (x) - sin (2 x) cos (2 x) = 0

x = πn, x = \frac{π}{2} + πn

Solution steps

2 sin (x) cos (x) - sin (2 x) cos (2 x) = 0

Rewrite using trig identities

sin (2 x) - cos (2 x) sin (2 x) = 0

Factor

sin (2 x) - cos (2 x) sin (2 x) : - sin (2 x) (cos (2 x) - 1)

- sin (2 x) (cos (2 x) - 1) = 0

Solving each part separately

sin (2 x) = 0 or cos (2 x) - 1 = 0

sin (2 x) = 0 : x = πn, x = \frac{π}{2} + πn

cos (2 x) - 1 = 0 : x = πn

Combine all the solutions

x = πn, x = \frac{π}{2} + πn

cos (x) = \frac{21}{29}

\frac{4}{sin ( 104.4 8 ^{\circ} )} = \frac{2}{sin ( a )}

sin (x + \frac{3 π}{2}) - cos (x + \frac{π}{2}) = 1

csc (x) - cot (x) = 1

sin (x) = - \frac{1}{2} sin (2 x)

What is the general solution for 2sin(x)cos(x)-sin(2x)cos(2x)=0 ?
The general solution for 2sin(x)cos(x)-sin(2x)cos(2x)=0 is x=pin,x= pi/2+pin