Chương 1: HÀM SỐ LƯỢNG GIÁC. PHƯƠNG TRÌNH LƯỢNG GIÁC

Hàm xác định trên R khi với mọi x ta có:

\(sin^6x+cos^6x+m.sinx.cosx>0\)

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\dfrac{m}{2}sin2x>0\)

\(\Leftrightarrow1-\dfrac{3}{4}sin^22x+\dfrac{m}{2}sin2x>0\)

Đặt \(sin2x=t\in\left[-1;1\right]\)

\(\Rightarrow-3t^2+2m.t+4>0\)

\(\Leftrightarrow\min\limits_{\left[-1;1\right]}f\left(t\right)>0\)

Xét hàm \(f\left(t\right)=-3t^2+2mt+4\) có \(a=-3< 0\) ; \(-\dfrac{b}{2a}=\dfrac{m}{3}\) ; 

 \(f\left(-1\right)=1-2m\) ; \(f\left(1\right)=1+2m\)

TH1: \(\dfrac{m}{3}\le-1\Rightarrow m\le-3\)

\(\Rightarrow\min\limits_{\left[-1;1\right]}f\left(t\right)=f\left(1\right)=1+2m>0\Rightarrow m>-\dfrac{1}{2}\) (ktm)

TH2: \(\dfrac{m}{3}\ge1\Rightarrow m\ge3\)

\(\Rightarrow\min\limits_{\left[-1;1\right]}f\left(t\right)=f\left(-1\right)=1-2m>0\Rightarrow m< \dfrac{1}{2}\) (ktm)

TH3: \(-1< \dfrac{m}{3}< 1\Rightarrow-3< m< 3\)

\(\Rightarrow\min\limits_{\left[-1;1\right]}f\left(t\right)=min\left\{1-2m;1+2m\right\}\)

TH3.1: \(\left\{{}\begin{matrix}1-2m\le1+2m\\1-2m>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\ge0\\m< \dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow0\le m< \dfrac{1}{2}\)

TH3.2: \(\left\{{}\begin{matrix}1+2m\le1-2m\\1+2m>0\end{matrix}\right.\) \(\Rightarrow-\dfrac{1}{2}< m\le0\)

Kết hợp lại ta được: \(-\dfrac{1}{2}< m< \dfrac{1}{2}\)

\(\left(sinx-1\right)\left(cosx+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)

\(4sin^2x-2\left(\sqrt{3}+\sqrt{2}\right)sinx+\sqrt{6}=0\)

\(\Leftrightarrow4sin^2x-2\sqrt{3}sinx-2\sqrt{2}sinx+\sqrt{6}=0\)

\(\Leftrightarrow2sinx\left(2sinx-\sqrt{3}\right)-\sqrt{2}\left(2sinx-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left(2sinx-\sqrt{3}\right)\left(2sinx-\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{\sqrt{3}}{2}\\sinx=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\\x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

Chọn C

Chọn D

\(\Leftrightarrow1-cos2x-\left(1-cos^22x\right)=m.cos2x+m+2\)

\(\Leftrightarrow cos^22x-cos2x-2-m\left(cos2x+1\right)=0\)

\(\Leftrightarrow\left(cos2x+1\right)\left(cos2x-2\right)-m\left(cos2x+1\right)=0\)

\(\Leftrightarrow\left(cos2x+1\right)\left(cos2x-m-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=m+2\end{matrix}\right.\)

Pt \(cos2x=-1\) có 1 nghiệm \(x=\dfrac{\pi}{2}\) thuộc đoạn đã cho nên pt có đúng 2  nghiệm thuộc đoạn đã cho khi \(cos2x=m+2\) có đúng 1 nghiệm khác \(\dfrac{\pi}{2}\) trên đoạn đã cho

\(\Rightarrow-\dfrac{1}{2}< m+2\le1\Rightarrow-\dfrac{5}{2}< m\le-1\)

\(\Rightarrow m=\left\{-2;-1\right\}\) có 2 giá trị nguyên

Áp dụng BĐT \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\):

\(y^2=\left(\sqrt{sinx}+\sqrt{1-sinx}\right)^2\le sinx+1-sinx=1\)

\(\Rightarrow-1\le y\le1\)

\(\Rightarrow M^4-m^4=0\)