# challenge set for solving trig equations

• September 3rd 2010, 05:45 AM
aeroflix
challenge set for solving trig equations
• September 3rd 2010, 05:48 AM
Prove It
a) Cross multiply, expand, move everything to one side, factorise and solve.

b) Multiply both sides by $\tan{x}$ to get $\tan^2{x} = 1$. Solve.
• September 3rd 2010, 07:24 AM
Soroban
Hello, aeroflix!

Quote:

$(3)\;\;\cot^{\text{-}1}(2x) - \tan^{\text{-}1}(x) \:=\:30^o$

Note that: . $\cot^{\text{-}1}(2x) \;=\;\tan^{\text{-}1}\!\left(\frac{1}{2x}\right)$ . . . (Think about it!)

So we have: . $\tan^{\text{-}1}\!\left(\frac{1}{2x}\right) - \tan^{\text{-}1}(x) \:=\:30^o$

Take the tangent of both sides:

. . $\tan\bigg[\tan^{\text{-}1}\!\left(\frac{1}{2x}\right) - \tan^{\text{-}1}(x)\bigg] \;=\;\tan30^o$

. . $\dfrac{\tan\left[\tan^{\text{-}1}(\frac{1}{2x})\right] - \tan\left[\tan^{\text{-}1}(x)\right]} {1 + \tan\left[\tan^{\text{-}1}(\frac{1}{2x})\right] \tan\left[\tan^{\text{-}1}(x)\right]} \;=\;\dfrac{1}{\sqrt{3}}$

. . $\dfrac{\frac{1}{2x} - x}{1 - (\frac{1}{2x})(x)} \;=\;\dfrac{1}{\sqrt{3}} \quad\Rightarrow\quad 2\sqrt{3}\,x^2 + 3x - \sqrt{3} \:=\:0$

Quadratic Formula: . $x \;=\;\dfrac{-3 \pm\sqrt{33}}{4\sqrt{3}}$

Therefore: . $x \;=\;\begin{Bmatrix}\;\;0.396143496 \\ \text{-}1.262168899 \end{Bmatrix}$
• September 3rd 2010, 02:33 PM
aeroflix
yes i remember it,, its the same step just like the tan 60 degrres before right???? how did u know and figure out how to solve it?? whats your trick?
• September 3rd 2010, 09:09 PM
Prove It
Experience
• September 4th 2010, 08:16 AM
aeroflix
here,, i solved letter b but its wrong

i ended up with 1/2 is not equal to 1/3

heres my solution or scratchhttp://img186.imageshack.us/img186/1323/qqqqqqqqqqy.png
• September 4th 2010, 08:22 AM
e^(i*pi)
Quote:

Originally Posted by aeroflix
here,, i solved letter b but its wrong

i ended up with 1/2 is not equal to 1/3

heres my solution or scratchhttp://img186.imageshack.us/img186/1323/qqqqqqqqqqy.png

You're trying too hard. Recall that $\cot(x) = \frac{1}{\tan(x)}$

Thus $\tan(x) = \frac{1}{\tan(x)}$

Multiply by $\tan(x)$

$\tan^2(x) = 1$ which is equal to $\tan^2(x)-1^2 = 0$

Solve the quadratic, I would use the difference of two squares to solve

Edit: Oops, I posted the answer to B not A which is in the quote

From what I can tell you're right (although confusing lol) up to $\sec^2(x) =2$. However, when you took the square root you didn't take the negative.

$\sec(x) = \pm \sqrt{2}$

Taking the reciprocal of both sides: $\cos(x) = \pm \frac{1}{\sqrt{2}}$

I got answers of $\frac{\pi}{4}$ and $-\frac{\pi}{4}$ (or $\pm 45^{\circ}$ ) as solutions (although only principal solutions because no range is defined in the question)

$\sec(x) \neq 0$ so discard that solution (from factoring out $\sec(x)$)
• September 4th 2010, 08:27 AM
Prove It
From $2\sec{x} + \sec^2{x} = \sec^2{x} + \sec^3{x}$

$2\sec{x} = \sec^3{x}$

$0 = \sec^3{x} - 2\sec{x}$

$0 = \sec{x}(\sec^2{x} - 2)$

$0 = \sec{x}(\tan^2{x} - 1)$ since $\tan^2{x} = \sec^2{x} - 1$

$\sec{x} = 0$ or $\tan^2{x} - 1 = 0$

Case 1: Does $\sec{x}$ ever equal $0$?

Case 2: $\tan^2{x} - 1 = 0$

$\tan^2{x} = 1$

$\tan{x} = \pm 1$

$x = \frac{\pi}{4} + \frac{n\pi}{2}$ where $n \in \mathbf{Z}$

$x = \frac{\pi + 2n\pi}{4}$ where $n \in \mathbf{Z}$

$x = \frac{(2n + 1)\pi}{4}$ where $n \in \mathbf{Z}$.
• September 4th 2010, 05:23 PM
aeroflix
where did n pi / 2 came from?
• September 4th 2010, 05:34 PM
aeroflix
what is the porper way t o solve sec^2x = sqaure root of 2