Tan x + sqrt(3) = - 2 tan x

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  nπ -π/6 . . . for any integer n

Step-by-step explanation:

  tan(x) +√3 = -2tan(x) . . . . . given

  3tan(x) = -√3 . . . . . . . . . . . add 2tan(x)-√3

  tan(x) = -√3/3 . . . . . . . . . . divide by 3

  x = arctan(-√3/3) = -π/6 . . . . use the inverse tangent function to find x

This is the value in the range (-π/2, π/2). The tangent function repeats with period π, so the set of values of x that will satisfy this equation is ...

  x = n·π -π/6 . . . . for any integer n

sss (side, side, side)

sas (side, angle, side)

asa (angle, side, angle)

aas (angle, angle, side)

hl (hypotenuse, leg)