1. ## 2sinx-cotxcosx=1

the problem is...

$2sin\Theta-cot\Theta cos\Theta=1$

after switching terms around and doing some operations i got...

$4sin^2\Theta=1+\dfrac{2}{sin^3\Theta}+\dfrac{1}{si n^6\Theta}$

This feels like the wrong direction since i'm using terms with sin^6 in the denominator.

When i first looked at it, it seemed like getting all the terms to be sin (or cos too i guess) would be possible. Well i got all the terms to be in sin, but these terms have large exponents in the denominator, etc.

Is there another, easier approach?

thanks!

2. Originally Posted by jonnygill
the problem is...

$2sin\Theta-cot\Theta cos\Theta=1$

-Dan

3. Originally Posted by topsquark
$\displaystyle 2 sin( \theta ) - cot( \theta) cos( \theta ) = 0$

$\displaystyle 2 sin( \theta ) - \frac{cos^2( \theta)}{sin(\theta )} = 0$

$\displaystyle 2 sin^2( \theta ) - cos^2( \theta) = 0$

Now use $\displaystyle sin^2( \theta ) = 1 - cos^2( \theta )$

-Dan

could you show how you went from line 2 to line 3?

i tried getting a common denominator but i ended up with $-\dfrac{1}{sin^3\Theta}+2sin\Theta$

also, where did the 1 go in the original equation? why did you just remove it?

edit: i see how you went from line 2 to line 3. but i do not understand why you set the equation equal to zero and not one. was this an oversight?

4. i kept the one and got $3sin^2\Theta-sin\Theta-1=0$

I can now use the quad. formula to solve for sin(Theta). I suspect you removed the 1 on purpose. perhaps it makes it easier to solve. but how would you re-incorporate that 1 back into the equation when you are finished?

edit: i solved the equation using the quad. formula and got two answers that work with the original equation. Thanks!

5. Originally Posted by jonnygill
the problem is...

$2sin\Theta-cot\Theta cos\Theta=1$

after switching terms around and doing some operations i got...

$4sin^2\Theta=1+\dfrac{2}{sin^3\Theta}+\dfrac{1}{si n^6\Theta}$

This feels like the wrong direction since i'm using terms with sin^6 in the denominator.

When i first looked at it, it seemed like getting all the terms to be sin (or cos too i guess) would be possible. Well i got all the terms to be in sin, but these terms have large exponents in the denominator, etc.

Is there another, easier approach?

thanks!
Simply use

$\displaystyle\frac{cos\Theta}{sin\Theta}=cot\Theta$

which gives

$\displaystyle\ 2sin\Theta-\frac{cos\Theta}{sin\Theta}\;cos\Theta=1\Rightarro w\left[\frac{sin\Theta}{sin\Theta}\right]2sin\Theta-\frac{cos^2\Theta}{sin\Theta}=1$

which gives a common denominator

$\displaystyle\frac{2sin^2\Theta-cos^2\Theta}{sin\Theta}=1$

$2sin^2\Theta-\left(1-sin^2\Theta\right)=sin\Theta$