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Solve 2cos^2x-sinx-1=0 on the interval [0,2pi)

 Jul 21, 2021
 #1
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I have not answered this but I would start by substituting   

1) sin^2x - 1 = cos^2x

2) then I'd tidy the equation

3) substitute y for sinx

4) solve as a quadratic equation

5) solve sinx= your solutions.

 Jul 22, 2021

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