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Can someone help me solve csc^2 x = 5 algebraically over the interval 0 <= x < 2pi?
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So far, I have sin x = 1/sqrt5
when we take square root we get 2 solutions \[\sqrt{x^2}= \pm x \]
Yes
and since we cant leave the radical sign at the denominator we should multiply top and bottom of the fraction by the denominator
So 1/sqrt5 is basically sqrt5/5
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x = sin^-1 sqrt5/5
yes right now i think we should use calculator to find exact values but here is sign chat that helps us to figure out the quadrant |dw:1445737916263:dw|
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