revving81
revving81 revving81
  • 18-03-2017
  • Mathematics
contestada

Solve the equation on the interval [0,2π). cos^4x=cos^4xcscx

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LammettHash
LammettHash LammettHash
  • 19-03-2017
[tex]\cos^4x=\cos^4x\csc x[/tex]
[tex]\cos^4x-\cos^4x\csc x=0[/tex]
[tex]\cos^4x(1-\csc x)=0[/tex]
[tex]\implies \begin{cases}\cos^4x=0\\1-\csc x=0\end{cases}\implies x=\dfrac\pi2,\dfrac{3\pi}2[/tex]
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