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

The general solution for tan(2x)+tan(4x)=0 is x=(pin)/3 ,x= pi/6+(pin)/3

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

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

x = \frac{πn}{3}, x = \frac{π}{6} + \frac{πn}{3}

Solution steps

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

Express with sin, cos

\frac{cos ( 2 x ) sin ( 4 x ) + cos ( 4 x ) sin ( 2 x )}{cos ( 2 x ) cos ( 4 x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Rewrite using trig identities

sin (2 x + 4 x) = 0

General solutions for

sin (2 x + 4 x) = 0

2 x + 4 x = 0 + 2 πn, 2 x + 4 x = π + 2 πn

Solve

2 x + 4 x = 0 + 2 πn : x = \frac{πn}{3}

Solve

2 x + 4 x = π + 2 πn : x = \frac{π}{6} + \frac{πn}{3}

x = \frac{πn}{3}, x = \frac{π}{6} + \frac{πn}{3}

2 sin (θ) - 3 = 0, 0 \leq θ < 2 π

arctan (θ) = - 1

tan^{2} (θ) - tan (θ) - 2 = 0

sin (x) = cos^{2} (x) - 1

tan (x) = a, tan (x + 18 0^{\circ})

What is the general solution for tan(2x)+tan(4x)=0 ?
The general solution for tan(2x)+tan(4x)=0 is x=(pin)/3 ,x= pi/6+(pin)/3