What is the general solution for sin(4x)+2sin(2x)=0 ?

The general solution for sin(4x)+2sin(2x)=0 is x=pin,x=(pi+2pin)/2

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

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

x = πn, x = \frac{π + 2 πn}{2}

Solution steps

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

Let:

u = 2 x

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

Rewrite using trig identities

2 sin (u) + 2 cos (u) sin (u) = 0

Factor

2 sin (u) + 2 cos (u) sin (u) : 2 sin (u) (cos (u) + 1)

2 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}

cos (2 x) = - \frac{1}{2}

2 sec^{2} (x) + 5 sec (x) + 2 = 0

sin^{2} (θ) = 8 (cos (- θ) + 1)

tan (x) = \frac{7.1}{5}

cos^{2} (x) + 3 sin (x) - 3 = 0

What is the general solution for sin(4x)+2sin(2x)=0 ?
The general solution for sin(4x)+2sin(2x)=0 is x=pin,x=(pi+2pin)/2