What is the general solution for 5sin(4x)=-10sin(2x) ?

The general solution for 5sin(4x)=-10sin(2x) is x=pin,x=(pi+2pin)/2

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5sin(4x)=-10sin(2x)

5 sin (4 x) = - 10 sin (2 x)

x = πn, x = \frac{π + 2 πn}{2}

Solution steps

5 sin (4 x) = - 10 sin (2 x)

Subtract

- 10 sin (2 x)

from both sides

5 sin (4 x) + 10 sin (2 x) = 0

Let:

u = 2 x

5 sin (2 u) + 10 sin (u) = 0

Rewrite using trig identities

10 sin (u) + 10 cos (u) sin (u) = 0

Factor

10 sin (u) + 10 cos (u) sin (u) : 10 sin (u) (cos (u) + 1)

10 sin (u) (cos (u) + 1) = 0

Solving each part separately

sin (u) = 0 or cos (u) + 1 = 0

sin (u) = 0 : u = 2 πn, u = π + 2 πn

cos (u) + 1 = 0 : u = π + 2 πn

Combine all the solutions

u = 2 πn, u = π + 2 πn

Substitute back

u = 2 x

2 x = 2 πn : x = πn

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

x = πn, x = \frac{π + 2 πn}{2}

cot^{4} (x) = cot^{4} (x) sin (x)

2 - csc^{2} (x) = 0

tan (x) = \frac{- 3}{3}

2 arctanh (\frac{x - 2}{x + 1}) = ln (2)

sec (β) = 3, \frac{3 π}{2} < β < 2 π

What is the general solution for 5sin(4x)=-10sin(2x) ?
The general solution for 5sin(4x)=-10sin(2x) is x=pin,x=(pi+2pin)/2