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timurjin [86]
3 years ago
9

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

Mathematics
2 answers:
pantera1 [17]3 years ago
7 0

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}}}

garik1379 [7]3 years ago
6 0

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} \\

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