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

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

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sin(2x)cos(x)=cos(2x)sin(x)

sin (2 x) cos (x) = cos (2 x) sin (x)

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (2 x) cos (x) = cos (2 x) sin (x)

Subtract

cos (2 x) sin (x)

from both sides

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

Rewrite using trig identities

sin (2 x) cos (x) - cos (2 x) sin (x)

Use the Angle Difference identity:

sin (s) cos (t) - cos (s) sin (t) = sin (s - t)

= sin (2 x - x)

sin (2 x - x) = 0

General solutions for

sin (2 x - x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x - x = 0 + 2 πn, 2 x - x = π + 2 πn

2 x - x = 0 + 2 πn, 2 x - x = π + 2 πn

Solve

2 x - x = 0 + 2 πn : x = 2 πn

2 x - x = 0 + 2 πn

Add similar elements:

2 x - x = x

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

Solve

2 x - x = π + 2 πn : x = π + 2 πn

2 x - x = π + 2 πn

Add similar elements:

2 x - x = x

x = π + 2 πn

x = 2 πn, x = π + 2 πn

sin (x) = \frac{7}{12}

sin (x) = \frac{7}{13}

3 sin (x) - sin (x) - 3 = 0

cos (x) + 2 cos (x + 24 0^{\circ}) = 1

tan (x) + 2 sin (x) = 0

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