if -180<=x<=180 solve 6-10cosx=3sin^2x
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Originally Posted by rionariona if -180<=x<=180 solve 6-10cosx=3sin^2x $\displaystyle \sin^2(x) = 1 - \cos^2(x)$ $\displaystyle 6-10 \cos(x) = 3 - 3\cos^2(x)$ which is a quadratic in $\displaystyle cos(x)$ Put into the form $\displaystyle ax^2+bx+c=0$ and solve using the quadratic formula
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