# Thread: Trig. equation solution problem.

1. ## Trig. equation solution problem.

tan(x) + 3cot(x) = 5sec(x)

so far I have arrived at

(tan(x))2 = (0.3334)1/2
tan(x) = 0.5774 & -0.5774
x = 300 & -300

However the solution states x = 300 as the principal solution, only. Which is correct since when x = -300 is substituted into the original equation the LHS and RHS of the equation
are numerically equal but opposite in sign.

That said my question is, what am I missing? How am I supposed to ascertain which is the correct value of x (as the square root of a number has both -ve and +ve solutions) without substituting x back into the original equation. Or is it expected that i do exactly that, substitute back into the equation for all solutions derived directly from the square root of number.

Ooh one other thing what do you guys use to input mathematical equations or formulea more efficiently on this board?

Thank You.

2. ## Re: Trig. equation solution problem.

Originally Posted by Gayelle

tan(x) + 3cot(x) = 5sec(x)

so far I have arrived at

(tan(x))2 = (0.3334)1/2
tan(x) = 0.5774 & -0.5774
x = 300 & -300

However the solution states x = 300 as the principal solution, only. Which is correct since when x = -300 is substituted into the original equation the LHS and RHS of the equation
are numerically equal but opposite in sign.

That said my question is, what am I missing? How am I supposed to ascertain which is the correct value of x (as the square root of a number has both -ve and +ve solutions) without substituting x back into the original equation. Or is it expected that i do exactly that, substitute back into the equation for all solutions derived directly from the square root of number.

Ooh one other thing what do you guys use to input mathematical equations or formulea more efficiently on this board?

Thank You.
I'd probably do it this way

\displaystyle \begin{align*} \tan{x} + 3\cot{x} &= 5\sec{x} \\ \frac{\sin{x}}{\cos{x}} + \frac{3\cos{x}}{\sin{x}} &= \frac{5}{\cos{x}} \\ \sin{x} + \frac{3\cos^2{x}}{\sin{x}} &= 5 \\ \sin{x} + \frac{3\left(1 - \sin^2{x}\right)}{\sin{x}} &= 5 \\ \sin^2{x} + 3\left(1 - \sin^2{x}\right) &= 5\sin{x} \\ \sin^2{x} + 3 - 3\sin^2{x} &= 5\sin{x} \\ 0 &= 2\sin^2{x} + 5\sin{x} - 3 \\ 0 &= 2\sin^2{x} + 6\sin{x} - \sin{x} - 3 \\ 0 &= 2\sin{x}\left(\sin{x} + 3\right) - 1\left(\sin{x} + 3\right) \\ 0 &= \left(\sin{x} + 3\right)\left(2\sin{x} - 1\right) \\ \sin{x} + 3 = 0 \textrm{ or } 2\sin{x} - 1 &= 0 \\ 2\sin{x} - 1 &= 0 \textrm{ (we can't use }\sin{x} + 3 = 0\textrm{ because }|\sin{x}| \leq 1 \textrm{ for all }x \textrm{)} \\ 2\sin{x} &= 1 \\ \sin{x} &= \frac{1}{2} \\ x &= \left\{ \frac{\pi}{3}, \pi - \frac{\pi}{3} \right\} + 2\pi n \textrm{ where } n \in \mathbf{Z} \\ x &= \left\{\frac{\pi}{3}, \frac{2\pi}{3}\right\} + 2\pi n \end{align*}

3. ## Re: Trig. equation solution problem.

Daaaamn that was fast!!!

First I would like to thank you "Prove it". So, THANK YOU.

However although I thoroughly understand what you just explained, and appreciate it a whole lot, I have one question which might sound stupid. Why wont it work my way? I've always thought that once
I make the correct substitutions it matters not what route I take in maths we should all arrive at the same result. Was I incorrect in substituting sec(x) = (1 + (tan(x))^2)^(1/2)
to solve the equation and if not how do i determine which substitutions are best.

Also I was wandering how you guys are able to state your questions and answers so orderly and efficiently as you just did. Please keep in mind that i am a complete noobie to this whole board thing and when i am entering my mathematical statements i feel as though i am attempting eye surgery with a blunt axe.

Thanks again man.

4. ## Re: Trig. equation solution problem.

The problem is that \displaystyle \begin{align*} -30^{\circ} \end{align*} is NOT a solution.

Try substituting it into your equation...

\displaystyle \begin{align*} LHS &= \tan{\left(-30^{\circ}\right)} + 3\cot{\left(-30^{\circ}\right)} \\ &= -\frac{\sqrt{3}}{3} - 3\sqrt{3} \\ &= \frac{-10\sqrt{3}}{3} \\ \\ RHS &= 5\sec{\left(-30^{\circ}\right)} \\ &= 5\left(\frac{2}{\sqrt{3}}\right) \\ &= \frac{10\sqrt{3}}{3} \\ &\neq LHS \end{align*}

Which makes me wonder what substitutions you actually used, because there must be a mistake...

5. ## Re: Trig. equation solution problem.

Nothing wrong with your substitution. We were always taught that if one step of a solution involves squaring both sides and a later step is to take square root of both sides then always check whether your answers fit the original equation.
A simple example would be (x+5)=3 (x+5)^2=9 So x+5=3 or -3 So x=-2 or -8. But of course only x=-2 fits the original.

6. ## Re: Trig. equation solution problem.

I'm a little embarrassed to put up my entire working as i took the loooong
way round (your way is much simpler). However i would still like to know why my way did not work. So here i go please look it over.

tanx + 3cotx = 5secx
tanx + 3cotx = 5(1 + tan2x)1/2 since sec2x = 1 + tan2x
X + (3/X) = 5(1 +X2)1/2 let tanx = X
X/5 + 3/(5X) = (1 +X2)1/2
(X/5 + 3/(5X))2 = (1 +X2)
X2/25 + 2(X/5)(3/(5X)) + 9/(25X2) = (1 +X2)
X2 + 6 + 9/(X2) - 25 -25X2 = 0
24X4 + 19X2 - 9 = 0 multiplying throughout by -X2
24Y2 + 19Y -9 = 0 let Y = X2
Y2 + 2(19/48)Y + (19/48)2 - (19/48)2 - (9/24) = 0 completing the square
Y = -1.12503 & 0.333367
X2 = 0.333367 since Y = X2 also no such thing as sqrt of -ve number.
X = 0.5774 & -0.5774 tanx = X
tanx = 0.5774 & -0.5774
x = 300 & -300

Thanks

7. ## Re: Trig. equation solution problem.

So it was that I must simply resubmit my results into the equation on completion to insure the correct answer.

Thanks biffboy.

8. ## Re: Trig. equation solution problem.

Originally Posted by Gayelle
So it was that I must simply resubmit my results into the equation on completion to insure the correct answer.

Thanks biffboy.
Pretty much. What Biffboy is saying is that squaring nearly always brings in extraneous solutions.

9. ## Re: Trig. equation solution problem.

Got it. Thanks Prove It