SOLUTION: find all solutions: cscx + cotx = 1 ; in the interval [0,2pi)

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Question 254929: find all solutions:
cscx + cotx = 1 ; in the interval [0,2pi)
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
find all solutions:
csc%28x%29+%2B+cot%28x%29+=+1 ; in the interval [0,2pi)
first
csc(x) = 1/sin(x)
cot(x) = cos(x) / sin(x)
so we get
1%2Fsin%28x%29+%2B+cos%28x%29+%2F+sin%28x%29+=+1
multiply by sin(x) to get
1+%2B+cos%28x%29+=+sin%28x%29
sin%5E2%28x%29+=+1+-+cos%5E2%28x%29+=+%281%2Bcos%28x%29%29%281-cos%28x%29%29
substituting, we get
1+%2B+cos%28x%29+=+sqrt%281%2Bcos%28x%29%29%281-cos%28x%29%29
squaring boh sides, we get
%281+%2B+cos%28x%29%29%281%2Bcos%28x%29%29+=+%281%2Bcos%28x%29%29%281-cos%28x%29%29
dividing by (1+cos(x)), we get
1+%2B+cos%28x%29+=+1-cos%28x%29
cos%28x%29+=+0
x+=+pi%2F2
x+=+3pi%2F2
--
If x = pi/2,
then
csc(x) = 1 and cot(x) = 0; 1 + 0 = 1
if x = 3pi/2,
then
csc(x) = -1 and cot(x) = 0 ; -1 + 0 not = 1.
So it appear that our only answer is
X = PI/2.