How do you solve $secxcscx=2cscx$ and find all solutions in the interval $[0,2pi)$ ?

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Answer 1 · 2016-07-14T12:17:43

$cos x = pi/3 and cos x = (5pi)/3$

=> $1/cos x * 1/ sin x = 2 1/sin x$
=> $1/cos x * cancel(1/ sin x) = 2 cancel(1/sin x )$
=> $cos x = 1/2$
=> $cos x = cos (pi/3)$ ...[Ans]

cos is positive at $I$ quadrant and $IV$ quadrant
so for $IV$ quadrant, angle will be $(2pi - theta)$

$cos x = cos (2pi-theta)$

$cos x = cos (2pi-(pi/3))$

$cos x = cos ((5pi)/3)$ ... [Ans]

How do you solve secxcscx=2cscxsecxcscx=2cscx and find all solutions in the interval [0,2pi)[0,2pi)?