bakerbw5 bakerbw5
  • 11-05-2018
  • Mathematics
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what are solutions to the equation sinx cosx = square root 3 over4

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sqdancefan
sqdancefan sqdancefan
  • 12-05-2018
[tex]\sin(x)\cos(x)=\dfrac{\sqrt{3}}{4}\\\\2\sin(x)\cos(x)=\dfrac{\sqrt{3}}{2}\\\\\sin(2x)=\dfrac{\sqrt{3}}{2}\\\\2x=\arcsin\left( \dfrac{\sqrt{3}}{2} \right)\\\\2x=\left\{ \dfrac{\pi}{3},\dfrac{2\pi}{3}\right\}+2n\pi\\\\\boxed{x=\left\{ \dfrac{\pi}{6},\dfrac{\pi}{3}\right\}+n\pi}[/tex]
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