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- Sep 3rd 2010, 04:45 AMaeroflixchallenge set for solving trig equations
- Sep 3rd 2010, 04:48 AMProve It
a) Cross multiply, expand, move everything to one side, factorise and solve.

b) Multiply both sides by $\displaystyle \tan{x}$ to get $\displaystyle \tan^2{x} = 1$. Solve. - Sep 3rd 2010, 06:24 AMSoroban
Hello, aeroflix!

Quote:

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

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

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

Take the tangent of both sides:

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

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

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

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

Therefore: . $\displaystyle x \;=\;\begin{Bmatrix}\;\;0.396143496 \\ \text{-}1.262168899 \end{Bmatrix}$

- Sep 3rd 2010, 01:33 PMaeroflix
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?

- Sep 3rd 2010, 08:09 PMProve It
Experience

- Sep 4th 2010, 07:16 AMaeroflix
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 - Sep 4th 2010, 07:22 AMe^(i*pi)
You're trying too hard. Recall that $\displaystyle \cot(x) = \frac{1}{\tan(x)}$

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

Multiply by $\displaystyle \tan(x)$

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

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 $\displaystyle \sec^2(x) =2$. However, when you took the square root you didn't take the negative.

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

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

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

$\displaystyle \sec(x) \neq 0$ so discard that solution (from factoring out $\displaystyle \sec(x)$) - Sep 4th 2010, 07:27 AMProve It
From $\displaystyle 2\sec{x} + \sec^2{x} = \sec^2{x} + \sec^3{x}$

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

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

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

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

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

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

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

$\displaystyle \tan^2{x} = 1$

$\displaystyle \tan{x} = \pm 1$

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

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

$\displaystyle x = \frac{(2n + 1)\pi}{4}$ where $\displaystyle n \in \mathbf{Z}$. - Sep 4th 2010, 04:23 PMaeroflix
where did n pi / 2 came from?

- Sep 4th 2010, 04:34 PMaeroflix
what is the porper way t o solve sec^2x = sqaure root of 2