# Thread: a few questions from homework

1. ## a few questions from homework

I have to find all answers to these on the interval 0-2 pi, I think I'm getting them right but I'm not 100% positive

For the first one I had
$\displaystyle cos x-2sinxcosx=0$

and I did
$\displaystyle (2cosx-2sinxcosx)/cosx=0$
$\displaystyle 1-2sinx=0$
$\displaystyle 1=2sinx$
$\displaystyle sinx=1/2$

and on the interval 0-2pi I got
$\displaystyle pi/6, 5pi/6$

the second one I have to do
$\displaystyle 2sin^2 x=4sinx+6$

and I got
$\displaystyle sin^2 x=2sinx+3$
$\displaystyle sin^2 x-2sinx-3=0$
$\displaystyle x^2-2x-3=0$
$\displaystyle (x-3)(x+1)$
$\displaystyle sinx=3 sinx=-1$

and for the interval I got
$\displaystyle 3pi/2$

are there any answers on the interval that I am missing?

also if anyone can help me with this identity
(cscx-cotx)^2=(1-cosx)/(1+cosx)

I've tried everything but have gotten nowhere

2. Originally Posted by prerunner
I have to find all answers to these on the interval 0-2 pi, I think I'm getting them right but I'm not 100% positive

For the first one I had
$\displaystyle cos x-2sinxcosx=0$

and I did
$\displaystyle (2cosx-2sinxcosx)/cosx=0$ Mr F says: Wrong. You have to factorise:

$\displaystyle {\color{red}\cos x (1 - 2 \sin x) = 0 \Rightarrow \cos x = 0}$ or $\displaystyle {\color{red}1 - 2 \sin x = 0}$.

$\displaystyle 1-2sinx=0$
$\displaystyle 1=2sinx$
$\displaystyle sinx=1/2$

and on the interval 0-2pi I got
$\displaystyle pi/6, 5pi/6$ Mr F says: For reasons given above, you're missing solutions.

the second one I have to do
$\displaystyle 2sin^2 x=4sinx+6$

and I got
$\displaystyle sin^2 x=2sinx+3$
$\displaystyle sin^2 x-2sinx-3=0$
$\displaystyle x^2-2x-3=0$ Mr F advises: Use a different symbol to x. x is already being used in the question. Use w = sin x so that w^2 - 2w - 3 = 0.

$\displaystyle (x-3)(x+1)$
$\displaystyle sinx=3 sinx=-1$

and for the interval I got
$\displaystyle 3pi/2$

are there any answers on the interval that I am missing? Mr F says: No.

[snip]
..

3. Originally Posted by prerunner
[snip]
also if anyone can help me with this identity
(cscx-cotx)^2=(1-cosx)/(1+cosx)

I've tried everything but have gotten nowhere
Left Hand Side $\displaystyle = (\text{cosec} \, x - \text{cot}\, x)^2$

$\displaystyle = \left(\frac{1}{\sin x} - \frac{\cos x}{\sin x}\right)^2$

$\displaystyle = \left(\frac{1 - \cos x}{\sin x}\right)^2$

$\displaystyle = \frac{(1 - \cos x)^2}{\sin^2 x}$

$\displaystyle = \frac{(1 - \cos x)^2}{1 - \cos^2 x}$

$\displaystyle = \frac{(1 - \cos x)^2}{(1 - \cos x)(1 + \cos x)}$

= ....