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

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

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

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

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

Solution steps

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

Subtract

- cos (3 x) + sin (4 x)

from both sides

sin (2 x) + cos (3 x) - sin (4 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 + \frac{π}{2}) = - cos (x)

sin (x) = (\frac{3}{5})

x, sin^{2} (x) = cos^{2} (x)

cos (θ) = 1 + 2 cos (θ)

cos (β) = \frac{2}{7}

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