Question

[kim anh tran]

Tìm GTLN, GTNN của hàm số y = 3- cos2x - 3sinx.

Answer 1

[Sonboygaming Tran]

cos2x=1-2sin²x

y=3+2sin²x-1-3sinx

y=2sin²x-3sinx+2

y=2(sin²x-\(\dfrac{3}{2}\)x+1)

y=2.(sin²x-2.1.\(\dfrac{3}{4}\).sinx+\(\dfrac{9}{16}\)+\(\dfrac{7}{16}\))

y=2.[sin²x-2.1.\(\dfrac{3}{4}\).sinx+(\(\dfrac{3}{4}\))² ]+\(\dfrac{7}{8}\)

y=2.(sinx-\(\dfrac{3}{4}\))²+\(\dfrac{7}{8}\)

Ta có:

-1\(\le\)sinx\(\le\)1

\(\dfrac{-7}{4}\)\(\le\)sinx-\(\dfrac{3}{4}\)\(\le\)1/4

0\(\le\)(sinx-\(\dfrac{3}{4}\))²\(\le\)1/16

0\(\le\)2(sinx-\(\dfrac{3}{4}\))²\(\le\)1/8

7/8\(\le\)2(sinx-\(\dfrac{3}{4}\))²+7/8\(\le\)1

7/8\(\le\)y\(\le\)1

=>min_y=7/8<=>sinx-3/4=0<=>\(\left\{{}\begin{matrix}x=arcsin\dfrac{3}{4}+k2\Pi\\x=\Pi-arcsin\dfrac{3}{4}+k2\Pi\end{matrix}\right.\)

max_y=1<=>sinx=1<=>x=\(\dfrac{\Pi}{2}\)+k2\(\Pi\)

Answer 2

[ĐỖ THỊ NGỌC ÁNH]

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