How do you find the critical numbers of $y=cos x + sin 2x$ ?

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Answer 1 · 2018-01-13T19:24:55

$y'=-sinx+2cos2x = -sinx+2(1-2sin^2x)$

$y' = -4sin^2x-sinx+2$

$y' = 0$ at solutions to

$4sin^2x+sinx-2=0$

$sinx = (-1+-sqrt(1+32))/8 = (-1 +-sqrt33)/8$

$x= {(sin^-1(-1-sqrt33)/8 +2pik),(-pi-sin^-1(-1-sqrt33)/8 +2pik) ,(sin^-1(-1+sqrt33)/8 +2pik),(pi-sin^-1(-1+sqrt33)/8 +2pik):}$ $" "$ for integer $k$

How do you find the critical numbers of y=cos x + sin 2xy=cos x + sin 2x?