find all the solutions from 0 to 2pi 2cos^2x = 13sinx-5
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Originally Posted by jumpman23 find all the solutions from 0 to 2pi 2cos^2x = 13sinx-5 $\displaystyle \cos^2{x} = 1 - \sin^2{x}$ ... solve the resulting quadratic for $\displaystyle \sin{x}$ btw ... why are these basic trig problems on the calculus page?
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