anonymous
  • anonymous
5sin^2theta-4sintheta+cos2theta=0
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Remove cos^2 theta using 1-sin^2 theta, then\[5\sin^2 \theta - 4 \sin \theta +1 - \sin ^ 2 \theta =0\]\[=4\sin^2 \theta - 4 \sin \theta +1 =0\]The quadratic equation then gives,\[\sin \theta = \frac{1}{2}\](there's a double root). The general solution is then\[\theta = n.180^o+(-1)^n \sin ^{-1}\frac{1}{2}\]i.e.\[\theta = n.180^o+(-1)^n 30^o\]
anonymous
  • anonymous
for n any integer.
anonymous
  • anonymous
so what would the answer be in radians?

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anonymous
  • anonymous
Well 180 degrees is equivalent to pi radians, and 30 degrees is (30/180)pi = pi/6 radians.
anonymous
  • anonymous
\[\theta = n \pi +(-1)^n \frac{\pi}{6}\]
anonymous
  • anonymous
You're welcome.

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