What is the general solution for sin(4x)-cos(3x)=sin(2x) ?

The general solution for sin(4x)-cos(3x)=sin(2x) is x= pi/6+(2pin)/3 ,x= pi/2+(2pin)/3

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

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

x = \frac{π}{6} + \frac{2 πn}{3}, x = \frac{π}{2} + \frac{2 πn}{3}

Solution steps

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

Subtract

sin (2 x)

from both sides

sin (4 x) - cos (3 x) - sin (2 x) = 0

Rewrite using trig identities

- cos (3 x) + 2 cos (3 x) sin (x) = 0

Factor

- cos (3 x) + 2 cos (3 x) sin (x) : cos (3 x) (2 sin (x) - 1)

cos (3 x) (2 sin (x) - 1) = 0

Solving each part separately

cos (3 x) = 0 or 2 sin (x) - 1 = 0

cos (3 x) = 0 : x = \frac{π}{6} + \frac{2 πn}{3}, x = \frac{π}{2} + \frac{2 πn}{3}

2 sin (x) - 1 = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + \frac{2 πn}{3}, x = \frac{π}{2} + \frac{2 πn}{3}, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Merge Overlapping Intervals

x = \frac{π}{6} + \frac{2 πn}{3}, x = \frac{π}{2} + \frac{2 πn}{3}

sin (x) cos (x) = \frac{1}{2} tan (x)

x, 2 sec^{2} (x) + tan^{2} (x) - 3 = 0

5 sin (\frac{θ}{2}) = 5 cos (\frac{θ}{2})

tan (x) = \frac{80}{50}

- cos (x) + 4 cos (2 x) = 0

What is the general solution for sin(4x)-cos(3x)=sin(2x) ?
The general solution for sin(4x)-cos(3x)=sin(2x) is x= pi/6+(2pin)/3 ,x= pi/2+(2pin)/3