Question #978fc

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Answer 1 · 2018-02-12T10:10:09

#theta=pi(2n+1/6)# with #ninZZ#

We want to find #theta# in

#csc(theta)=2#

Take the reciprocal on both sides

#sin(theta)=1/2#

By the inverse sine

#theta=pi(2n+1/6), pi(2n+5/6)# with #ninZZ#

Notice #sin(theta)>0# in the 1. and 2. quadrant,

and #cos(theta)<0# in the 2. and 3. quadrant

The only #theta#, which satisfy these criteria* must be

#theta=pi(2n+5/6)# with #ninZZ#

/

*Notice #theta=pi(2n+1/6)# falls in the 1. quadrant,
and therefore don't satisfy #cos(theta)<0#

While #theta=pi(2n+5/6)# falls in the 2. quadrant,
and therefore satisfy both criteria