# precalculus

2cosxcscsx-4cosx-cscx+2=0

solutions?

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1. Factorize by grouping of terms:

2cos(x)csc(x)-4cos(x)-csc(x)+2 = 0

csc(x)(2cos(x)-1) -2(2cos(x)-1) = 0
(csc(x)-2)(2cos(x)-1) = 0
Therefore
csc(x)-2 = 0 OR 2cos(x)-1 = 0
If csc(x)-2=0, 1/sin(x)=2, sin(x)=1/2
If 2cos(x)-1=0, cos(x)=1/2

The question does not mention if it will require a general solution or all solutions between 0 and 2π. So find the respective solutions accordingly.

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2. Easy

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