e8lleyfingsarappl
e8lleyfingsarappl e8lleyfingsarappl
  • 16-07-2016
  • Mathematics
contestada

solve cos^2(2x)-sin^2(2x)=sqrt3/2

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dalendrk
dalendrk dalendrk
  • 25-07-2016
[tex]\cos^22x-\sin^22x=\dfrac{\sqrt3}{2}\\\\\cos(4x)=\dfrac{\sqrt3}{2}\iff4x=30^o\to x=7.5^o\\----------------------\\\cos^22x-\sin^22x=\dfrac{\sqrt3}{2}\\\\\cos(4x)=\dfrac{\sqrt3}{2}\iff4x=\dfrac{\pi}{6}+2k\pi\ or\ 4x=-\dfrac{\pi}{6}+2k\pi\\\\\boxed{x=\dfrac{\pi}{24}+\dfrac{k\pi}{2}\ or\ x=-\dfrac{\pi}{24}+\dfrac{k\pi}{2}\ for\ k\in\mathbb{Z}}[/tex]
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