LidiaVoelz LidiaVoelz
  • 25-09-2015
  • Mathematics
contestada

cos2x + sin( x - (5pi/2)) +1=0

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Julik
Julik Julik
  • 25-09-2015
[tex]cos(2x)+sin(x- \frac{5 \pi }{2} )+1=0 \\ \\ cos(2x)=2cos^2(x)-1 \\ sin(x- \frac{5 \pi }{2} )= -cos(x) \\ \\ 2cos^2(x)-1 -cos(x) +1=0 \\ 2cos^2(x) -cos(x)=0 \\ cos(x)(2cos(x)-1)=0 \\ \\ cos(x)=0 \\ x= \frac{ \pi }{2} + \pi k,~~~k \in Z \\ [/tex]

[tex]\\ 2cos(x)-1=0 \\ 2cos(x)=1 \\ cos(x)= \frac{1}{2} \\ x= \pm \frac{ \pi }{3} +2 \pi k,~~k \in Z[/tex]
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