Need assistance with 2 problems

[hey_eddie]

Hi,
I have a take home test and I need help getting started on these 2 problems:

1. Find the exact values of x in the interval [0, 2pi):
2 sin[sup:10x4231l]2[/sup:10x4231l] x cot x cot x = 0

I set everything equal to zero, but I end up with a + or - neg square root and 1/2 which is impossible.

2. Verify the identity:

sin u = 1 + sec u - tan u
csc u


Any help would be appreciated!!

 

[Loren]

>2 sin[sup:2k040f9p]2[/sup:2k040f9p] x cot x cot x = 0

If no typo, then cot x times cot x = cot[sup:2k040f9p]2[/sup:2k040f9p]x. From there, change cot to cos/sin and go.

For number 2 change all terms into terms of sin and cos. On the right side transform into three fractions. The rest is duck soup.

 

[hey_eddie]

So sorry. There was a typo:

Find the exact values of x in the interval [0, 2 pi)

2 sin[sup:2pqt1uua]2[/sup:2pqt1uua]x cot x + cot x = 0

So sorry.

 

[soroban]

Hellp, hey_eddie!

$\displaystyle \text{1. Find the exact values of }x\text{ in the interval }[0,\:2\pi):\!\quad 2\sin^2\!x\cot x + \cot x \:=\: 0$

$\displaystyle \text{Factor: }\;\cot x\left(2\sin^2\!x + 1\right) \:=\:0$

\(\displaystyle \text{And we have: }\;\left\{\begin{array}{cccccccccc}2\sin^2\!x + 1 \;=\; 0 & \Rightarrow & \sin^2\!x \:=\:-\frac{1}{2} & ?? \\ \\[-3mm] \cot x \;=\; 0 & \Rightarrow &\boxed{ x \;=\; \frac{\pi}{2},\:\frac{3\pi}{2}} \end{array}\)

$\displaystyle \text{2. Verify the identity: }\;\sin x \;=\;\frac{1+\sec x}{\csc x} - \tan x$

$\displaystyle \text{The right side is: }\;\frac{1 + \frac{1}{\cos x}}{\frac{1}{\sin x}} - \tan x$

$\displaystyle \text{Multiply the fraction by }\frac{\sin x}{\sin x}\!:\quad \frac{\sin x\left(1 + \frac{1}{\cos x}\right)}{\sin x(\csc x)} \;-\; \tan x \;=\;\frac{\sin x + \frac{\sin x}{\cos x}}{1} \;-\; \tan x$

. . $\displaystyle = \;\;\sin x + \tan x - \tan x \;\;=\;\;\sin x$