Trig expressions - Simplify

[SnickersFrampus]

Studying for a Trig test tomorrow. Having trouble with:

Instructions say to Simplify:

1. cot x tan 2x = 3

2. csc x sin 2x = 2 + tan (1/2) x

 

[barson90]

Try swapping parts out with the right identities.
http://www.sosmath.com/trig/Trig5/trig5/trig5.html

 

[fasteddie65]

1. cot x tan 2x = 3
1/tan x • 2 tan x/(1 - tan^2 x) = 2/(1 - tan^2 x) = 3
1 - tan^2 x = 2/3
1/3 = tan^2 x
±1/?3 = tan x
x = ±?/6, ±5?/6, ...

 

[soroban]

Hello, SnickersFrampus!

\(\displaystyle \text{I will solve for the interval }\,[0,\:2\pi)\)

Instructions say to Simplify
. . You mean "Solve for x", don't you?

\(\displaystyle (1)\;\;\cot x \tan 2x \:=\: 3\)

\(\displaystyle \text{We use the identity: }\:\tan2A \:=\:\frac{2\tan A}{1 - \tan^2\!A}\)


\(\displaystyle \text{We have: }\;\frac{1}{\tan x}\cdot\frac{2\tan x}{1 - \tan^2\!x} \:=\:3 \quad\Rightarrow\quad \frac{2}{1-\tan^2\!x} \:=\:3 \quad\Rightarrow\quad 2 \:=\:3-3\tan^2\!x\)

. . . . . . . .\(\displaystyle 3\tan^2\!x \:=\:1 \quad\Rightarrow\quad \tan^2\!x \:=\:\frac{1}{3} \quad\Rightarrow\quad \tan x \:=\:\pm\frac{1}{\sqrt{3}}\)

\(\displaystyle \text{Therefore: }\;x \;=\;\frac{\pi}{6},\:\frac{5\pi}{6},\:\frac{7\pi}{6},\:\frac{11\pi}{6}\)

\(\displaystyle (2)\;\;\csc x \sin 2x = 2 + \tan\frac{x}{2}\)

\(\displaystyle \text{We use these identities: }\;\begin{array}{ccc}\sin 2A &=& 2\si A\cos A \\\\[-3mm] \tan\frac{A}{2} &=& \dfrac{1-\cos A}{\sin A} \end{array}\)


\(\displaystyle \text{We have: }\;\frac{1}{\sin x}\cdot 2\sin x\cos x \;=\;2 + \frac{1-\cos x}{\sin x}\)

\(\displaystyle \text{Multiply by }\sin x\!:\;\;2\sin x\cos x \:=\:2\sin x \,+\, 1 \,-\, \cos x \quad\Rightarrow\quad 2\sin x\cos x \,+\, \cos x \,-\, 2\sin x \,-\, 1 \:=\:0\)

\(\displaystyle \text{Factor: }\;\cos x(2\sin x + 1) - (2\sin x + 1) \:=\:0 \quad\Rightarrow\quad (2\sin x + 1)(\cos x-1) \:-\:0\)


\(\displaystyle \text{And we have:}\)

. . \(\displaystyle \begin{array}{ccccccccc}2\sin x+1\:=\:0 & \Rightarrow & \sin x \:=\:-\frac{1}{2} & \Rightarrow & x \:=\:\frac{7\pi}{6},\:\frac{11\pi}{6} \\ \\[-3mm]\cos x -1\:=\:0 & \Rightarrow & \cos x \:=\:1 & \Rightarrow\quad & x \:=\:0 \end{array}\)


\(\displaystyle \text{But }x = 0\text{ is an extraneous root.}\)

\(\displaystyle \text{Therefore: }\;x \;=\;\frac{7\pi}{6},\:\frac{11\pi}{6}\)