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Tìm m để hàm số : \(y=\sqrt{\frac{m-\sin x-\cos x-2\sin x\cos x}{\sin^{2017}x-\cos^{2019}x \sqrt{2}}}\) xác định với mọi... - Hoc24

Câu 1:

\(cos7x-\sqrt{3}sin7x=-2\\ \Leftrightarrow cos\left(7x+\dfrac{\pi}{3}\right)=-1\\ \Leftrightarrow7x+\dfrac{\pi}{3}=-\pi+k2\pi\\ \Leftrightarrow x=-\dfrac{4\pi}{21}+k\dfrac{2\pi}{7}\)

Vì \(x\in[\dfrac{2\pi}{5};\dfrac{6\pi}{7}]\)

\(\Rightarrow\dfrac{2\pi}{5}\le x\le\dfrac{6\pi}{7}\\ \Leftrightarrow\dfrac{2\pi}{5}\le-\dfrac{4\pi}{21}+k\dfrac{2\pi}{7}\le\dfrac{6\pi}{7}\\ \Leftrightarrow\dfrac{31}{15}\le k\le\dfrac{11}{3}\)

Vì \(k\in Z\) nên \(k=3\)

Vậy \(x\) cần tìm là \(\dfrac{2\pi}{3}\)

Câu 2:

\(2sin^2x-sinxcosx-cos^2x=m\\ \Leftrightarrow2\dfrac{1-cos2x}{2}-\dfrac{1}{2}s\text{in2}x-\dfrac{1+cos2x}{2}=m\\ \Leftrightarrow3cos2x+s\text{in2}x=1-2m\)

Điều kiện để phương trình có nghiệm là:

\(3^2+1^2\ge\left(1-2m\right)^2\\ \Leftrightarrow4m^2-4m-9\le0\\ \Leftrightarrow\dfrac{1-\sqrt{10}}{2}\le m\le\dfrac{1+\sqrt{10}}{2}\)

\(\left\{{}\begin{matrix}\left|sinx\right|\le1\\\left|cosx\right|\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}sin^{4034}x\le sin^2x\\cos^{4038}x\le cos^2x\end{matrix}\right.\)

\(\Rightarrow sin^{4034}x+cos^{4038}x< sin^2x+cos^2x=1\) (dấu = ko xảy ra)

\(\Rightarrow\left|sin^{2017}x-cos^{2019}x\right|< \sqrt{\left(1+1\right)\left(sin^2x+cos^2x\right)}=\sqrt{2}\)

\(\Rightarrow sin^{2017}x-cos^{2019}x+\sqrt{2}>0\) \(\forall x\)

Vậy để hàm số xác định với mọi x trên đoạn đã cho

\(\Rightarrow m-sinx-cosx-2sinx.cosx\ge0\) \(\forall x\)

\(\Leftrightarrow sinx+cosx+2sinx.cosx\le m\)

Đặt \(sinx+cosx=t\Rightarrow2sinx.cosx=t^2-1\) \(\left(-1\le t\le\sqrt{2}\right)\)

\(\Rightarrow t^2+t-1\le m\Rightarrow m\ge\max\limits_{\left[-1;\sqrt{2}\right]}\left(t^2+t-1\right)=\sqrt{2}+1\)

Vậy \(m\ge\sqrt{2}+1\)

a/

\(\Leftrightarrow sin\left(x+\frac{\pi}{8}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{8}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{8}=\frac{5\pi}{6}+l2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{24}+k2\pi\\x=\frac{17\pi}{24}+l2\pi\end{matrix}\right.\)

\(\left\{{}\begin{matrix}-\pi\le\frac{\pi}{24}+k2\pi\le\pi\\-\pi\le\frac{17\pi}{24}+l2\pi\le\pi\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}k=0\\l=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{24}\\x=\frac{17\pi}{24}\end{matrix}\right.\) \(\Rightarrow\sum x=\frac{\pi}{24}+\frac{17\pi}{24}=\frac{3\pi}{4}\)

2.

\(4sin^22x-1=0\Leftrightarrow2-2cos4x-1=0\)

\(\Leftrightarrow cos4x=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{2}\\x=-\frac{\pi}{12}+\frac{l\pi}{2}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}-\frac{\pi}{2}\le\frac{\pi}{12}+\frac{k\pi}{2}\le\frac{\pi}{2}\\-\frac{\pi}{2}\le-\frac{\pi}{12}+\frac{l\pi}{2}\le\frac{\pi}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}k=\left\{-1;0\right\}\\l=\left\{0;1\right\}\end{matrix}\right.\)

\(\Rightarrow x=\left\{-\frac{5\pi}{12};\frac{\pi}{12};-\frac{\pi}{12};\frac{5\pi}{12}\right\}\Rightarrow\sum x=0\)

Câu 1 với câu 2 sai đề, sin và cos nằm trong [-1;1], mà căn 2 với căn 3 lớn hơn 1 rồi

3/ \(\sin x=\cos2x=\sin\left(\frac{\pi}{2}-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}-2x+k2\pi\\x=\pi-\frac{\pi}{2}+2x+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\frac{2}{3}\pi\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

4/ \(\Leftrightarrow\cos^2x-2\sin x\cos x=0\)

Xét \(\cos x=0\) là nghiệm của pt \(\Rightarrow x=\frac{\pi}{2}+k\pi\)

\(\cos x\ne0\Rightarrow1-2\tan x=0\Leftrightarrow\tan x=\frac{1}{2}\Rightarrow x=...\)

5/ \(\Leftrightarrow\sin\left(2x+1\right)=-\cos\left(3x-1\right)=\cos\left(\pi-3x+1\right)=\sin\left(\frac{\pi}{2}-\pi+3x-1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\frac{\pi}{2}-\pi+3x-1\\2x+1=\pi-\frac{\pi}{2}+\pi-3x+1\end{matrix}\right.\Leftrightarrow....\)

6/ \(\Leftrightarrow\cos\left(\pi\left(x-\frac{1}{3}\right)\right)=\frac{1}{2}\Leftrightarrow\pi\left(x-\frac{1}{3}\right)=\pm\frac{\pi}{3}+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{1}{3}+2k\Rightarrow x=\frac{2}{3}+2k\left(1\right)\\x-\frac{1}{3}=-\frac{1}{3}+2k\Rightarrow x=2k\left(2\right)\end{matrix}\right.\)

\(\left(1\right):-\pi< x< \pi\Rightarrow-\pi< \frac{2}{3}+2k< \pi\) (Ủa đề bài sai hay sao ý nhỉ?)

7/ \(\Leftrightarrow\left[{}\begin{matrix}5x+\frac{\pi}{3}=\frac{\pi}{2}-2x+\frac{\pi}{3}\\5x+\frac{\pi}{3}=\pi-\frac{\pi}{2}+2x-\frac{\pi}{3}\end{matrix}\right.\Leftrightarrow...\)

Thui, để đây bao giờ...hết lười thì làm tiếp :(

7)

\(sin\left(5x+\frac{\pi}{3}\right)=cos\left(2x-\frac{\pi}{3}\right)\)

\(\Leftrightarrow sin\left(5x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{2}-2x-\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+\frac{\pi}{3}=\frac{\pi}{2}-2x-\frac{\pi}{3}+k2\pi\\5x+\frac{\pi}{3}=\pi-\left(\frac{\pi}{2}-2x-\frac{\pi}{3}\right)+k2\pi\end{matrix}\right.\left(k\in Z\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-\pi}{42}+k\frac{2\pi}{7}\\x=\frac{\pi}{6}+k\frac{2\pi}{3}\end{matrix}\right.\left(k\in Z\right)\)

Do:\(0< x< \pi\)

\(Với:x=\frac{-\pi}{42}+k\frac{2\pi}{7}\left(k\in Z\right)\Rightarrow khôngtìmđượck\)

\(Với:x=\frac{\pi}{6}+k\frac{2\pi}{3}\left(k\in Z\right)\Leftrightarrow\frac{1}{4}< k< \frac{5}{4}\Rightarrow k=\left\{0;1\right\}\Rightarrow\left[{}\begin{matrix}k=0\Rightarrow x=\frac{\pi}{6}\\k=1\Rightarrow x=\frac{5\pi}{6}\end{matrix}\right.\)

Vậy nghiệm của pt là: \(x=\frac{\pi}{6};x=\frac{5\pi}{6}\)

8)

Biến đổi xong nó thành hàm nhìn gọn lắm :)

\(=\sin^6x+\cos^6x+\frac{3}{4}.\frac{1}{2}\left[\sin\left(2x+\frac{\pi}{4}-2x+\frac{\pi}{4}\right)+\sin\left(2x+\frac{\pi}{4}+2x-\frac{\pi}{4}\right)\right]\)

\(=1+\frac{3}{8}\left(\sin\frac{\pi}{4}+\sin4x\right)\)

Bạn biện luận nốt nhé

Bạn muốn bài cụ thể như nào? Giải tiếp hay giải thích kỹ các bước phân tích cho bạn?

a/

Đặt \(cosx=t\Rightarrow0< t\le1\)

\(\Rightarrow t^2-2mt+4\left(m-1\right)=0\)

\(\Leftrightarrow t^2-4-2m\left(t-2\right)=0\)

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

\(\Leftrightarrow t=2m-2\)

\(\Rightarrow0< 2m-2\le1\Rightarrow1< m\le\frac{3}{2}\)

b.

\(x\in\left(-\frac{\pi}{2};\frac{\pi}{2}\right)\Rightarrow\frac{x}{2}\in\left(-\frac{\pi}{4};\frac{\pi}{4}\right)\)

Đặt \(sin\frac{x}{2}=t\Rightarrow-\frac{\sqrt{2}}{2}< t< \frac{\sqrt{2}}{2}\)

\(\Rightarrow4t^2+2t+m-2=0\Leftrightarrow4t^2+2t-2=-m\)

Xét \(f\left(t\right)=4t^2+2t-2\) trên \(\left(-\frac{\sqrt{2}}{2};\frac{\sqrt{2}}{2}\right)\)

\(f\left(-\frac{\sqrt{2}}{2}\right)=-\sqrt{2}\) ; \(f\left(\frac{\sqrt{2}}{2}\right)=\sqrt{2}\) ; \(f\left(-\frac{1}{4}\right)=-\frac{9}{4}\)

\(\Rightarrow-\frac{9}{4}\le f\left(t\right)< \sqrt{2}\Rightarrow-\frac{9}{4}\le-m< \sqrt{2}\)

\(\Rightarrow-\sqrt{2}< m\le\frac{9}{4}\)