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anonymous
 5 years ago
5sin^2theta4sintheta+cos2theta=0
anonymous
 5 years ago
5sin^2theta4sintheta+cos2theta=0

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Remove cos^2 theta using 1sin^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
 5 years ago
Best ResponseYou've already chosen the best response.0so what would the answer be in radians?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well 180 degrees is equivalent to pi radians, and 30 degrees is (30/180)pi = pi/6 radians.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\theta = n \pi +(1)^n \frac{\pi}{6}\]
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