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solvefor x,-2sin(2x)-2sin(4x)=0

Solution

solve for

x, - 2 sin (2 x) - 2 sin (4 x) = 0

Solution

x = πn, x = \frac{π + 2 πn}{2}, x = \frac{π ( 3 n + 1 )}{3}, x = \frac{π ( 3 n + 2 )}{3}

+1

Solution steps

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

Let:

u = 2 x

- 2 sin (u) - 2 sin (2 u) = 0

Rewrite using trig identities

- 2 sin (u) - 4 cos (u) sin (u) = 0

Factor

- 2 sin (u) - 4 cos (u) sin (u) : - 2 sin (u) (2 cos (u) + 1)

- 2 sin (u) (2 cos (u) + 1) = 0

Solving each part separately

sin (u) = 0 or 2 cos (u) + 1 = 0

sin (u) = 0 : u = 2 πn, u = π + 2 πn

2 cos (u) + 1 = 0 : u = \frac{2 π}{3} + 2 πn, u = \frac{4 π}{3} + 2 πn

Combine all the solutions

u = 2 πn, u = π + 2 πn, u = \frac{2 π}{3} + 2 πn, u = \frac{4 π}{3} + 2 πn

Substitute back

u = 2 x

2 x = 2 πn : x = πn

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

2 x = \frac{2 π}{3} + 2 πn : x = \frac{π ( 3 n + 1 )}{3}

2 x = \frac{4 π}{3} + 2 πn : x = \frac{π ( 3 n + 2 )}{3}

x = πn, x = \frac{π + 2 πn}{2}, x = \frac{π ( 3 n + 1 )}{3}, x = \frac{π ( 3 n + 2 )}{3}

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