How do you solve $cos x cscx = 2cosx$ ?

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Answer 1 · 2016-04-01T12:25:01

For $ninZZ$

$x =2npi+pi/6 or x=2npi+(5pi)/6 or x = (2n+1)pi/2$

$cosx csc x = 2cosx$

$cosx 1/sinx=2cosx$

$cosx/sinx-2cosx=0$

$cosx(1/sinx-2)=0$

$cosx=0$ or $1/sinx-2=0$

$cosx=0 => x = (2n+1)pi/2, ninZZ$

or

$1/sinx=2 => sinx = 1/2$

$=> x =2npi+pi/6 or 2npi+(5pi)/6, ninZZ$

How do you solve cos x cscx = 2cosxcos x cscx = 2cosx?