Is sec3x=-2 unsolvable?

Question

So when doing this problem and dividing the -2 by 3 it comes out to secx=-2/3 which is not possible right? Because no other side can be greater than the hypotenuse. Maybe I don't know what im talking about whatsoever, but help would be greatly appreciated.

Answer 1

You cannot divide by a number inside any trig functions. Instead, try making it simpler. Since

 

SEC(x) =   1           your equation can be rewritten as:

              COS(X)

 

    1      = -2            and you can take a shortcut by flipping both sides:

COS(3X)

 

COS(3X) = - 1         COS(X) = 1/2 with a reference angle of π/3 and is negative in Quadrants II & III

                   2          In quadrant II, the true angle is π - π/3 = 2π/3

                               In quadrant III, the true angle is π + π/3 = 4π/3

If you want specific solutions, we'll continue from there. if you were asked for "general solutions" for all real numbers, add " + 2πK" to each equation. Either way, you have

3X = 2π/3         and 3X = 4π/3         and you meed only divide both sides by 3 to finish up. If you added the "k" part I mentioned, simply divide that by 3 as well.

 

Aside from your solution, I will note that you were right about SEC being unable to ever equal -2/3 or +2/3. Just as SIN and COS cannot wander beyond +/- 1, SEC and CSC cannot wander within that interval.