\(y=4\left(1-sin^2x\right)+2sinx+2=-4sin^2x+2sinx+6\)

Đặt \(sinx=t\in\left[-1;1\right]\Rightarrow y=f\left(t\right)=-4t^2+2t+6\)

\(-\dfrac{b}{2a}=\dfrac{1}{4}\in\left[-1;1\right]\)

\(f\left(-1\right)=0\) ; \(f\left(\dfrac{1}{4}\right)=\dfrac{25}{4}\); \(f\left(1\right)=4\)

\(\Rightarrow y_{max}=\dfrac{25}{4}\) khi \(sinx=\dfrac{1}{4}\)

\(y_{min}=0\) khi \(sinx=-1\)

Ta có: \(y=4cos^2x+2sinx+2=4-4sin^2x+2sinx+2=-4sin^2x+2sinx+6=-\left(4sin^2x-2sinx+\dfrac{1}{16}-\dfrac{1}{16}-6\right)=-\left(2sin^2x-\dfrac{1}{4}\right)^2+\dfrac{97}{16}\)

Ta có: \(-\left(2sin^2x-\dfrac{1}{4}\right)^2\le0\Rightarrow y\le\dfrac{97}{16}\)

Vậy \(y_{max}=\dfrac{97}{16}\)

cái này áp dụng hệ thức lượng thôi bạn

AH=căn 6^2-4,8^2=3,6cm

=>AC=6^2/3,6=10cm

Chọn C

Pt \(\Leftrightarrow\left(2sinx-1\right)\left(2sin2x-1\right)=3-4\left(1-sin^2x\right)\)

\(\Leftrightarrow2sin2x\left(2sinx-1\right)-2sinx+1=-1+4sin^2x\)

\(\Leftrightarrow2sin2x\left(2sinx-1\right)-\left(4sin^2x+2sinx-2\right)=0\)

\(\Leftrightarrow2sin2x\left(2sinx-1\right)-2\left(2sinx-1\right)\left(sinx+1\right)=0\)

\(\Leftrightarrow2\left(2sinx-1\right)\left(sin2x-sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\left(1\right)\\sin2x=sinx+1\left(2\right)\end{matrix}\right.\)

Từ (1) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\),\(k\in Z\)

Từ (2)\(\Leftrightarrow2sinx.cosx-sinx-1=0\)

(Cái này tạm thời nghĩ ko ra,tối làm :)

Có 1 nghiệm thôi nha bạn

Vì 3/1<>1/2

\(\Leftrightarrow\left(2sinx-1\right)\left(2sinx-1-sinx+\frac{3}{2}\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(sinx+\frac{1}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(f\left(1-3x\right)=2\left(1-3x\right)-\left(1-3x\right)^2=1-9x^2\)

A,C đúng