# 8sin^2(v)-2cos(v)=5

• Aug 29th 2006, 04:52 AM
a4swe
8sin^2(v)-2cos(v)=5
$8\sin(v)^2-2\cos(v)=5 \Leftrightarrow$
$8(1-\cos(v)^2)-2\cos(v)=5 \Leftrightarrow$
$8-8\cos(v)^2-2\cos(v)=5 \Leftrightarrow$
$1-\cos(v)^2-\frac{1}{4}\cos(v)=\frac{5}{8} \Leftrightarrow$
$\cos(v)^2+\frac{1}{4}\cos(v)-\frac{3}{8}=0 \Leftrightarrow$
$\cos(v)=\frac{-7+-\sqrt{7}}{4}$

The thing is that my book states that I can solve this one partly exact.
That is for half of the inifinite solutions.
But I can get nothing of this, no exact solutions at all.
Where am I wrong?

• Aug 29th 2006, 05:26 AM
Random333
Quote:

Originally Posted by a4swe
$8\sin(v)^2-2\cos(v)=5 \Leftrightarrow$
$8(1-\cos(v)^2)-2\cos(v)=5 \Leftrightarrow$
$8-8\cos(v)^2-2\cos(v)=5 \Leftrightarrow$
$1-\cos(v)^2-\frac{1}{4}\cos(v)=\frac{5}{8} \Leftrightarrow$
$\cos(v)^2+\frac{1}{4}\cos(v)-\frac{3}{8}=0 \Leftrightarrow$
$\cos(v)=\frac{-7+-\sqrt{7}}{4}$

The thing is that my book states that I can solve this one partly exact.
That is for half of the inifinite solutions.
But I can get nothing of this, no exact solutions at all.
Where am I wrong?

Seems to me it is like $8\sin^2(v)$ as opposed to $8\sin(v^2)$. I think you're worked that out though because your second last line agrees with my working pretty much.

$\cos(v)^2+\frac{1}{4}\cos(v)-\frac{3}{8}=0 \Leftrightarrow$
So, I just factorised here to give:

$(\cos(v)+\frac{3}{4})(\cos(v)-\frac{1}{2})=0$
And then solve

$\cos(v)=\frac{1}{2}$
$v=\frac{pi}{3}$ that's in radians obviously.

That's what I've figured from it but I'm not 100% sure if I'm right but it looks right to me.
• Aug 29th 2006, 05:49 AM
CaptainBlack
Quote:

Originally Posted by Random333
Seems to me it is like $8\sin^2(v)$ as opposed to $8\sin(v^2)$. I think you're worked that out though because your second last line agrees with my working pretty much.

$\cos(v)^2+\frac{1}{4}\cos(v)-\frac{3}{8}=0 \Leftrightarrow$
So, I just factorised here to give:

$(\cos(v)+\frac{3}{4})(\cos(v)-\frac{1}{2})=0$
And then solve

$\cos(v)=\frac{1}{2}$
$v=\frac{pi}{3}$ that's in radians obviously.

That's what I've figured from it but I'm not 100% sure if I'm right but it looks right to me.

Looks right to me to.

RonL
• Aug 29th 2006, 06:10 AM
a4swe
Quote:

Originally Posted by Random333
Seems to me it is like $8\sin^2(v)$ as opposed to $8\sin(v^2)$. I think you're worked that out though because your second last line agrees with my working pretty much.

Well, yes you are right it's my LaTeX skills that stoped me.
Next time I'll do it right.
Thank you.
• Aug 29th 2006, 06:38 AM
topsquark
Speaking of LaTeX, this will make things a little neater:
$\pm$ for your "+-".

-Dan
• Aug 29th 2006, 07:46 AM
Soroban
Hello, a4swe!

Am I missing something important?
. . I found all the solutions . . .
And I have no idea what your book means by "exact" solutions.

Quote:

$8\sin^2v - 2\cos v \:=\:5$

$8(1-\cos^2 v)-2\cos v \:=\:5$
$8-8\cos^2v - 2\cos v \:=\:5$

Why introduce fractions? . . . Don't you hate them as much as I do?

We have: . $8\cos^2v + 2\cos v - 3\:=\:0$

Factor: . $(2\cos v - 1)(4\cos v + 3)\:=\:0$

And solve:
. . $2\cos v - 1 \:=\:0\quad\Rightarrow\quad \cos v \,= \,\frac{1}{2}\quad\Rightarrow\quad v \,= \,\pm\frac{\pi}{3} + 2\pi n$
. . $4\cos v + 3 \:=\:0\quad\Rightarrow\quad \cos v$ $= \text{ -}\frac{3}{4}\quad\Rightarrow\quad v \:= \:\cos^{-1}\!\left(\text{-}\frac{3}{4}\right) + 2\pi n$

. . . and those are the exact solutions.

• Aug 29th 2006, 08:16 AM
CaptainBlack
Quote:

Originally Posted by Soroban
Hello, a4swe!

Am I missing something important?
. . I found all the solutions . . .
And I have no idea what your book means by "exact" solutions.

Why introduce fractions? . . . Don't you hate them as much as I do?

We have: . $8\cos^2v + 2\cos v - 3\:=\:0$

Factor: . $(2\cos v - 1)(4\cos v + 3)\:=\:0$

And solve:
. . $2\cos v - 1 \:=\:0\quad\Rightarrow\quad \cos v \,= \,\frac{1}{2}\quad\Rightarrow\quad v \,= \,\pm\frac{\pi}{3} + 2\pi n$
. . $4\cos v + 3 \:=\:0\quad\Rightarrow\quad \cos v$ $= \text{ -}\frac{3}{4}\quad\Rightarrow\quad v \:= \:\cos^{-1}\!\left(\text{-}\frac{3}{4}\right) + 2\pi n$

. . . and those are the exact solutions.

You know what it means :D

RonL