Asked by Hank
Find all the solutions from [0, 2pi]
cot^2x+csc=x
cot^2x+csc=x
Answers
Answered by
Anonymous
Well, you're not gong to make it easy, are you?
csc(?)
also, setting a trig function(x) = x is not amenable to solution by hand.
So, assuming you meant
cot^2(x)+csc(x) = 0,
since cot^2 = csc^2-1,
csc^2(x)-1+csc(x) = 0
csc(x) = (-1 +/- sqrt(5))/2
still not a usually used value.
csc(?)
also, setting a trig function(x) = x is not amenable to solution by hand.
So, assuming you meant
cot^2(x)+csc(x) = 0,
since cot^2 = csc^2-1,
csc^2(x)-1+csc(x) = 0
csc(x) = (-1 +/- sqrt(5))/2
still not a usually used value.
Answered by
Reiny
Just supposing you meant what you typed
cot^2x+csc=x
cos^2 x/sin^2 x + 1 sinx = x
wolfram shows 4 answers in your domain
http://www.wolframalpha.com/input/?i=solve+cot%5E2%28x%29+%2B+csc%28x%29+%3D+x
x = appr. 1.211, 2.343, 3.523 and 5.964
When substituted all are within an error of .01
cot^2x+csc=x
cos^2 x/sin^2 x + 1 sinx = x
wolfram shows 4 answers in your domain
http://www.wolframalpha.com/input/?i=solve+cot%5E2%28x%29+%2B+csc%28x%29+%3D+x
x = appr. 1.211, 2.343, 3.523 and 5.964
When substituted all are within an error of .01
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