Question

solve on the interval [0,2pi): 2cscx+5=1

Answer 1

$x\in[0,\ 2\pi)\\\\2\csc x+5=1\ \ \ \ |-5\\\\2\csc x=-4\ \ \ \ |:2\\\\\csc x=-2\\\\\dfrac{1}{\sin x}=-2\to\sin x=-\dfrac{1}{2}\to x=-\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{7\pi}{6}+2k\pi\\\\x\in[0,\ 2\pi)\to x=-\dfrac{\pi}{6}+2\pi=\dfrac{11\pi}{6}\ \vee\ x=\dfrac{7\pi}{6}\\\\Answer:\ \boxed{D.\ \dfrac{7\pi}{6},\ \dfrac{11\pi}{6}}}$

Answer 2

Answer: D. $\frac{7\pi }{6} ,\frac{11\pi }{6}$

Step-by-step explanation:

Isolating the function:

$2csc(x)+5=1\\2csc(x)=1-5\\csc(x)=\frac{-4}{2} \\csc(x)=-2$

Since $csc(x)=\frac{1}{sin(x)}$ then:

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

The angles in the interval [0,2$\pi$] that comply with this are $\frac{7\pi }{6} and \frac{11\pi }{6}$

$sin(\frac{7\pi }{6})=-\frac{1}{2} \\sin(\frac{11\pi }{6})=-\frac{1}{2}$