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cos2x+3sinx-1=0 , -pi<x<2pi (this is the range)

Quote: Originally Posted by jcfulton cos2x+3sinx-1=0 , -pi<x<2pi (this is the range) recall that $\displaystyle \cos 2x = 1 - 2 \sin^2 x$, so you have $\displaystyle 2 \sin^2 x -3 \sin x = 0$ which is another quadratic that you can deal with (hint: factor out sine)