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anonymous

  • one year ago

solve between 0 and 2pi cotxsin^2x=2cotx

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  1. jdoe0001
    • one year ago
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    \(\large { cot(x)sin^2(x)=2cot(x)\implies cot(x)sin^2(x)-2cot(x)=0 \\ \quad \\ cot(x)[sin^2(x)-2]=0\implies \begin{cases} cot(x)=0\\ sin^2(x)-2=0 \end{cases} }\) solve those for "x", using the inverse trigonometric function, or just using your Unit Circle

  2. Mertsj
    • one year ago
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    1. Subtract 2cotx from both sides 2. Factor out cotx 3. Set each factor equal to 0 and solve each equation.

  3. jdoe0001
    • one year ago
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    btw, 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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