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anonymous
 one year ago
solve between 0 and 2pi
cotxsin^2x=2cotx
anonymous
 one year ago
solve between 0 and 2pi cotxsin^2x=2cotx

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\large { cot(x)sin^2(x)=2cot(x)\implies cot(x)sin^2(x)2cot(x)=0 \\ \quad \\ cot(x)[sin^2(x)2]=0\implies \begin{cases} cot(x)=0\\ sin^2(x)2=0 \end{cases} }\) solve those for "x", using the inverse trigonometric function, or just using your Unit Circle

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.01. Subtract 2cotx from both sides 2. Factor out cotx 3. Set each factor equal to 0 and solve each equation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0btw, if you didn't get the lines from surjithayer on the previous posting you can just let us know that you dunno, or what part you don't get, and maybe just explain you'd repost to get more eyes
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