What is the general solution for tan(x)(tan(x)+1)=0 ?

The general solution for tan(x)(tan(x)+1)=0 is x=pin,x=(3pi)/4+pin

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tan(x)(tan(x)+1)=0

tan (x) (tan (x) + 1) = 0

x = πn, x = \frac{3 π}{4} + πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

tan (x) (tan (x) + 1) = 0

Solving each part separately

tan (x) = 0 or tan (x) + 1 = 0

tan (x) = 0 : x = πn

tan (x) = 0

General solutions for

tan (x) = 0

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

tan (x) + 1 = 0 : x = \frac{3 π}{4} + πn

tan (x) + 1 = 0

Move

1

to the right side

tan (x) + 1 = 0

Subtract

1

from both sides

tan (x) + 1 - 1 = 0 - 1

Simplify

tan (x) = - 1

tan (x) = - 1

General solutions for

tan (x) = - 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

x = πn, x = \frac{3 π}{4} + πn

cos (θ) = sin (2 θ)

θ, tan (θ) = 1

sec (x) csc (x) - 2 csc (x) = 0

csc (3 θ) = 6 sin (3 θ), 0 < θ < 2 π

3 tan (3 θ) - 1 = 0, 0 \leq θ \leq 2 π

What is the general solution for tan(x)(tan(x)+1)=0 ?
The general solution for tan(x)(tan(x)+1)=0 is x=pin,x=(3pi)/4+pin