anonymous
  • anonymous
Solve each of the following trigonometric equations for 0<=x<=360 a) cosec x= 9 sin x @ganeshie8
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
just to double check my answer
anonymous
  • anonymous
\[cosec~x=9~\sin~x\]\[\frac{ 1 }{ \sin~x }=9~\sin~x\]\[\sin^2~x=\frac{ 1 }{ 9 }\]\[\sin~x=\frac{ 1 }{ 3 }\]\[x=19.07,160.53\]
anonymous
  • anonymous
*19.47

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Nnesha
  • Nnesha
i have few mints i have to go somewhere but i don't how u did u get 160 but when we take square root \[\rm \sqrt{\sin^2x}=\sqrt{\frac{1}{9}} \rightarrow sinx= \pm \frac{1}{3}\] so take invs of sin sin^{-1}(-1/3) = -19.47 and use the fact circle add up to360 to get the positive angle add 360 with -19.47
anonymous
  • anonymous
oh i totally forget about the = and -... Thank you @Nnesha
anonymous
  • anonymous
*+
anonymous
  • anonymous
so the answer should be \[\sin~x=\frac{ 1 }{ 3 }~and~\sin~x=-\frac{ 1 }{ 3 }\]\[x=19.47,160.53~and~x=(180+19.47),(360-19.47)\]\[x=19.47,160.53,199.47,340.53\]am i correct? @Nnesha

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