Dalton asked in Science & MathematicsMathematics · 1 decade ago

# Solve: tan2x=3 over the interval [0,2pi]?

please show steps. thanks.

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• Puggy
Lv 7

tan(2x) = 3, on [0, 2pi]

Therefore

2x = arctan(3)

x = (1/2)arctan(3)

arctan fetches a solution from -pi/2 to pi/2, but since 3 is positive, it is fetched in quadrant 1. Since solutions are always pi apart, the two solutions are

x = { (1/2)arctan(3) , (1/2)arctan(3) + pi }

• 1 decade ago

One angle whose tangent is 3 is arctan(3). But remember that the trig functions are periodic. The tangent function has a period of pi (180 degrees). That means that all the angles whose tangent is 3 are arctan(3) + n*pi, where n is any integer.

So, taking the arctan of both sides of the orginal equation we get

2x = arctan(3) + n*pi

x = 1/2*arctan(3) +n/2 * pi