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

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

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

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

x = πn, x = \frac{π}{2} + πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (2 x) = u

2 u = u

2 u = u : u = 0

2 u = u

Move

u

to the left side

2 u = u

Subtract

u

from both sides

2 u - u = u - u

Simplify

u = 0

u = 0

Substitute back

u = sin (2 x)

sin (2 x) = 0

sin (2 x) = 0

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

sin (2 x) = 0

General solutions for

sin (2 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 = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn, 2 x = π + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

Combine all the solutions

x = πn, x = \frac{π}{2} + πn

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

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

cos (2 θ) = (\frac{57}{11})

sin (x) cos (x^{2}) = sin (x), 0 \leq x \leq 2 π

θ, \frac{cos ( θ )}{sin ( θ )} = - \frac{π}{6}

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