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anonymous

  • 4 years ago

solve 2secx+tanx=3 on the interval xE[0,2pi]

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  1. anonymous
    • 4 years ago
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    this is a pain.

  2. anonymous
    • 4 years ago
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    we have \[\frac{2}{\cos(x)}+\frac{\sin(x)}{\cos(x)}=\frac{2+\sin(x)}{\cos(x)}=3\] so \[2+\sin(x)=3\cos(x)\] \[3\cos(x)-\sin(x)=2\] \

  3. anonymous
    • 4 years ago
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    now we have to write \[3\cos(x)-\sin(x)\] as a single function of sine, we get something like \[\sqrt{10}\sin(x+\tan^{-1}(3))\]

  4. anonymous
    • 4 years ago
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    no that is a mistake

  5. anonymous
    • 4 years ago
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    \[-\sqrt{10}\sin(x-\tan^{-1}(3))=2\] i think it is right now

  6. anonymous
    • 4 years ago
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    i wonder if there is an easier way, but i don't see it. you can divide by \[-\sqrt{10}\]and take the inverse sine of the result

  7. anonymous
    • 4 years ago
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    let me try something else

  8. anonymous
    • 4 years ago
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    no just going around in circles. the above will do it. do you have another method you are supposed to use?

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