How do you solve $secxcscx=2cscx$ ?

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Answer 1 · 2016-10-15T09:28:26

General solution is $x=2npi+-pi/3$ , where $n$ is an integer.

$secxcscx=2cscx$

Hence, considering $cscx!=0$ , $secx=2$

and so $x=+-pi/3$ - note that $csc(+-pi/3)!=0$

and as all trigonometric functions are cyclic over $2pi$ ,

General solution is $x=2npi+-pi/3$ , where $n$ is an integer.

How do you solve secxcscx=2cscxsecxcscx=2cscx?