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

please show steps. thanks.

### 2 Answers

- PuggyLv 71 decade agoFavorite Answer
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

But this gives us all the solutions of x. We only want the ones on the interval [0,2pi]. So what values of n keep our answer in that interval? Just plug in and see. 1/2*arctan(3) = .624... radians. If we add 1/2 pi we get 2.195... radians. Still in range. Add another 1/2 pi (n=2) and we get 3.766... radians. Still in range. Add another 1/2 pi (n=3) and we get 5.336... radians. Still in range. Add another 1/2 pi (n=4) and we get 6.907... radians. Out of range (2pi = 6.283... radians). So our answer is:

1/2*arctan(3) + n/2 * pi, for n = 0, 1, 2, 3.