Trig Identities, even possible?

Question

I am trying to solve for this trig identity, but I just cant do it. I do different steps and I am getting different answers, but they are wrong. Here is the problem:

tan2XcotX=3

Solve for an identities angle measure. Best answers to the best explanation!

Thanks in advance.

Answer 1

You're mistaken; this isn't a trig identity; it's a trig equation. (Trig identities are ones which are always true).

tan(2x)cot(x) = 3

Using the identity
tan(2x) = 2tan(x)/[1 - tan^2(x)], we have

2tan(x)/[1 - tan^2(x)] [1/tan(x)]= 3

The tan(x) in the numerator cancels with the 1/tan(x).

2 / [1 - tan^2(x)] = 3

Multiplying both sides by [1 - tan^2(x)] gives us

2 = 3[1 - tan^2(x)]
2 = 3 - 3tan^2(x)

Moving the -3tan^2(x) to the left hand side and the 2 to the right hand side,

3tan^2(x) = 3 - 2
3tan^2(x) = 1
tan^2(x) = (1/3); therefore

tan(x) = +/- 1/sqrt(3)

The solutions to tan(x) = 1/sqrt(3), presuming a restriction of
0 <= x < 2pi, are x = {pi/6, 7pi/6}.

The solutions to tan(x) = -1/sqrt(3) are x = {5pi/6, 11pi/6}

Therefore, our solution set in the interval [0, 2pi) is:

x = {pi/6, 7pi/6, 5pi/6, 11pi/6}

In degrees, in the integral [0, 360), our solution set is

x = {30, 210, 150, 330}

Answer 2

The angle's measure within the first quadrant is 30 degrees.

[There are others related to that: 150, 210 and 330 degrees in the first four quadrants, or angles of [(180 n) +/- 30] degrees in general, for all integer ' n.']

I'll concentrate first on the 30 degree result. THAT approach will enable me to give a COMPLETELY SELF-CONTAINED DERIVATION, without any appeal to KNOWING or simply QUOTING the measure of the angle associated with a given value for tan x. (This means that the answer ultimately goes back to first principles.)

You can find the value of this basic first quadrant angle by using the tan (2 theta) formula:

tan (2x) = [2 tan (x)] / [1 - tan^2 (x)]. Then, since (tan x)(cot x) = 1,

tan (2x) cot (x) = 2 / [1 - tan^2 (x)] = 3. Therefore tan^2 (x) = 1 / 3, so

tan (x) = 1 / sqrt (3). [Note: tan (x) = - 1 / sqrt (3) is also possible.]

Now, 1 and sqrt (3) are two perpendicular sides of a (1, sqrt (3), 2) right-angled triangle. (You may sketch that triangle and recognize what the angle ' x ' must be.) Alternatively, that triangle is half of an equilateral triangle with all sides of length ' 2 ' and therefore equal inside angles of 60 degrees. Your angle ' x ' is half of one of those 60 degree angles, that is 30 degrees.

Other solutions follow by considering where in the other quadrants tan x = +/- [1 / sqrt (3)] is satisfied. (This can be done by considering a simple sketch of y = tan x versus x and noting that, with angles in degrees, tan (- x) = - tan (x), tan (180 - x) = tan (180 + x) = tan (x), and so on.) That leads to:

In total, within the usual four quadrants, the solution's angles are 30, 150, 210 and 330 degrees. But from the way that the question is posed, the angles are NOT limited to being just in the traditional four quadrants. Therefore we must further generalize the results, as I now do immediately below.

In general, then, ALL angles of the form [(180 n) +/- 30] degrees satisfy the equation, for arbitrary integer ' n.'

Live long and prosper.

Answer 3

tan2x = 2tanx/(1-tan^2x)
tan2xcotx = 2/(1-tan^2x) = 3
Solve for tanx,
tanx = ±1/√3
x = 30, 150, 210 and 330 degrees.

Answer 4

cosx/(a million-sinx) - secx = [cosx*(a million+sinx)]/[(a million-sinx)*(a million+sinx)] - secx = [cosx*(a million+sinx)]/[(cosx)^2] - secx = (a million + sinx)/cosx - secx = a million/cosx + sinx/cosx - secx = secx + tanx - secx = tanx