+0  
 
0
236
2
avatar+81 

Solve each equation on the interval \(0\leq\theta<2\pi\).

\(cot(\frac{2\theta}{3})=-\sqrt{3}\)

\(tan(\frac{\theta}{2}+\frac{\pi}{3})=1\)

\(cos(\frac{\theta}{3}-\frac{\pi}{4})=\frac{1}{2}\)

 Apr 22, 2019
 #1
avatar+7764 
0

\(0\le \theta < 2\pi\\ \implies0\le \dfrac{2\theta}{3} <\dfrac{4\pi}{3}\)

 

\(\cot \left(\dfrac{2\theta}{3}\right)= -\sqrt3\\ \tan\left(\dfrac{2\theta}{3}\right) = -\dfrac{1}{\sqrt3}\\ \dfrac{2\theta}{3} = \dfrac{5\pi}{6}\\ \theta = \dfrac{5\pi}{4}\)

.
 Apr 22, 2019
 #2
avatar+7764 
0

\(0\le\theta < 2\pi\\ \implies 0\le \dfrac{\theta}{2} < \pi\\ \implies \dfrac{\pi}{3} \le \dfrac{\theta}{2}+\dfrac{\pi}{3} < \dfrac{4\pi}{3}\)

 

\(\tan\left(\dfrac{\theta}{2}+\dfrac{\pi}{3}\right) = 1\\ \dfrac{\theta}{2}+\dfrac{\pi}{3} = \dfrac{5\pi}4\\ \theta = \dfrac{11\pi}{6}\)

.
 Apr 22, 2019

37 Online Users

avatar
avatar
avatar
avatar
avatar
avatar