What is the general solution for tan(x/2)tan(x)=-3 ?

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

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tan(x/2)tan(x)=-3

tan (\frac{x}{2}) tan (x) = - 3

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

Solution steps

tan (\frac{x}{2}) tan (x) = - 3

Subtract

- 3

from both sides

tan (\frac{x}{2}) tan (x) + 3 = 0

Let:

u = \frac{x}{2}

tan (u) tan (2 u) + 3 = 0

Rewrite using trig identities

3 + \frac{2 tan ^{2} ( u )}{1 - tan ^{2} ( u )} = 0

Solve by substitution

tan (u) = 3, tan (u) = - 3

tan (u) = 3 : u = \frac{π}{3} + πn

tan (u) = - 3 : u = \frac{2 π}{3} + πn

Combine all the solutions

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

Substitute back

u = \frac{x}{2}

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

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

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

sin (- x) = - \frac{1}{3}, tan (x) = - \frac{2}{4}

x, h = tan (x - \frac{π}{6})

sec (θ) + \frac{1}{cot ( θ )} = 1

(1 + 2 sin (2 θ))^{\frac{1}{3}} = 0

sin (\frac{x}{3}) cos (6 x) - cos (\frac{x}{3}) sin (6 x) = 0

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