Find the principal solution of the equation.cosec x = 2
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Answered by
11
We know that principal value branch of
![{cosec}^{ - 1} x \:\: is\:\: [-\frac{\pi}{2}, \frac{\pi}{2}] \\](https://tex.z-dn.net/?f=+%7Bcosec%7D%5E%7B+-+1%7D+x+%5C%3A%5C%3A+is%5C%3A%5C%3A+%5B-%5Cfrac%7B%5Cpi%7D%7B2%7D%2C+%5Cfrac%7B%5Cpi%7D%7B2%7D%5D+%5C%5C+ "{cosec}^{ - 1} x \:\: is\:\: [-\frac{\pi}{2}, \frac{\pi}{2}] \\")
-{0}
To solve the given equation we must keep consider that cosec and cosec inverse cancels each other only if x belongs to its principal value .
![cosec\: x =2\\ \\ x = {cosec}^{ - 1} (2)\\ \\ we \: know \: that \: cosec \: \frac{\pi}{6} = 2 \\\\ So\\\\ x= {cosec}^{ - 1} (cosec ( \frac{\pi}{6} )) \\ \\x= \frac{ \pi}{6} \: \: \: \: (x \: belongs \: to \: [- \frac{\pi}{2}, \frac{\pi}{2}])-(0)\\ \\ x = \frac{ \pi}{6} \\](https://tex.z-dn.net/?f=++cosec%5C%3A+x+%3D2%5C%5C+%5C%5C+x+%3D+%7Bcosec%7D%5E%7B+-+1%7D+%282%29%5C%5C+%5C%5C+we+%5C%3A+know+%5C%3A+that+%5C%3A+cosec+%5C%3A+%5Cfrac%7B%5Cpi%7D%7B6%7D+%3D+2+%5C%5C%5C%5C+So%5C%5C%5C%5C+x%3D+%7Bcosec%7D%5E%7B+-+1%7D+%28cosec+%28+%5Cfrac%7B%5Cpi%7D%7B6%7D+%29%29+%5C%5C+%5C%5Cx%3D++%5Cfrac%7B+%5Cpi%7D%7B6%7D+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%28x+%5C%3A+belongs+%5C%3A+to+%5C%3A+%5B-+%5Cfrac%7B%5Cpi%7D%7B2%7D%2C+%5Cfrac%7B%5Cpi%7D%7B2%7D%5D%29-%280%29%5C%5C+%5C%5C+x+%3D+%5Cfrac%7B++%5Cpi%7D%7B6%7D+%5C%5C+ "cosec\: x =2\\ \\ x = {cosec}^{ - 1} (2)\\ \\ we \: know \: that \: cosec \: \frac{\pi}{6} = 2 \\\\ So\\\\ x= {cosec}^{ - 1} (cosec ( \frac{\pi}{6} )) \\ \\x= \frac{ \pi}{6} \: \: \: \: (x \: belongs \: to \: [- \frac{\pi}{2}, \frac{\pi}{2}])-(0)\\ \\ x = \frac{ \pi}{6} \\")
is the solution of the equation.
To solve the given equation we must keep consider that cosec and cosec inverse cancels each other only if x belongs to its principal value .
is the solution of the equation.
Answered by
2
Pi/6 and 5pi/6
Step-by-step explanation:
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