π 3π nπ 2 f π f 3π 2 π nπ, 3π nπ, n - WebAssign

⇒ f′(x)=2cosx+2sinxcosx =0 ⇔ 2cosx(1+sinx)=0 ⇔ cosx =0 or sinx = −1, so x = π 2 +2nπ or 3π 2 +2nπ, where n is any integer. Now f π 2 = 3 and f 3π 2 = −1, so the points on the curve with a horizontal tangentare π 2 +2nπ,3 and 3π 2 +2nπ,−1,wheren isanyinteger. ................
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