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At what point on the curve y=x+2cos x is the tangent horizontal in the interval [-2pi, 2pi]

User Anirvan
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\bf y=x+2cos(x)\qquad \qquad [-2\pi ,2\pi ]\\\\ -----------------------------\\\\ \cfrac{dy}{dx}=1-2sin(x)\implies 0=1-2sin(x)\implies sin(x)=\cfrac{1}{2} \\\\\\ \measuredangle x=sin^(-1)\left( (1)/(2) \right)\implies \measuredangle x= \begin{cases} (\pi )/(6)\\\\ (5\pi )/(6) \end{cases}

only in the 1st and 2nd quadrants, where sine is positive
User Milovan Zogovic
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