What is the general solution for tan^2(x)+4tan(x)+1=0 ?

The general solution for tan^2(x)+4tan(x)+1=0 is x=-0.26179…+pin,x=-1.30899…+pin

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tan^2(x)+4tan(x)+1=0

tan^{2} (x) + 4 tan (x) + 1 = 0

x = - 0.26179 \dots + πn, x = - 1.30899 \dots + πn

Solution steps

tan^{2} (x) + 4 tan (x) + 1 = 0

Solve by substitution

tan (x) = - 2 + 3, tan (x) = - 2 - 3

tan (x) = - 2 + 3 : x = arctan (- 2 + 3) + πn

tan (x) = - 2 - 3 : x = arctan (- 2 - 3) + πn

Combine all the solutions

x = arctan (- 2 + 3) + πn, x = arctan (- 2 - 3) + πn

Show solutions in decimal form

x = - 0.26179 \dots + πn, x = - 1.30899 \dots + πn

sin (c) = \frac{3}{4}

2 cosh^{2} (x) - 5 sinh (x) = 5

10 + 5 cos (2 x) = 15 cos (x)

y, x = cos (2 y)

tan (x) = \frac{a}{b}

What is the general solution for tan^2(x)+4tan(x)+1=0 ?
The general solution for tan^2(x)+4tan(x)+1=0 is x=-0.26179…+pin,x=-1.30899…+pin