1) Cho \(3\sin^4x-cos^4x=\frac{1}{2}\), tìm giá trị của B=\(sin^4x+3cos^4x\)
2) Cho \(4sin^4x+3cos^4x=\frac{7}{4}\), tìm giá trị của C=\(3sin^4x+4cos^4x\)
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NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2019
\(3sin^4x+cos^4x=\dfrac{3}{4}\Leftrightarrow\dfrac{\left(sin^2x\right)^2}{1}+\dfrac{\left(cos^2x\right)^2}{3}=\dfrac{1}{4}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\dfrac{\left(sin^2x\right)^2}{1}+\dfrac{\left(cos^2x\right)^2}{3}\ge\dfrac{\left(sin^2x+cos^2x\right)^2}{1+3}=\dfrac{1}{4}\)
Dấu "=" xảy khi khi và chỉ khi: \(sin^2x=\dfrac{cos^2x}{3}\Rightarrow sin^4x=\dfrac{cos^4x}{9}\)
Thay vào biểu thức ban đầu:
\(3\left(\dfrac{cos^4x}{9}\right)+cos^4x=\dfrac{3}{4}\Leftrightarrow\dfrac{4}{3}cos^4x=\dfrac{3}{4}\Rightarrow cos^4x=\dfrac{9}{16}\)
\(\Rightarrow A=\dfrac{cos^4x}{9}+3cos^4x=\dfrac{9}{16.9}+\dfrac{3.9}{16}=\dfrac{7}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 12 2021
\(P=\sqrt{\left(1-cos^2x\right)^2+6cos^2x+3cos^4x}+\sqrt{\left(1-sin^2x\right)^2+6sin^2x+3sin^4x}\)
\(=\sqrt{4cos^4x+4cos^2x+1}+\sqrt{4sin^4x+4sin^2x+1}\)
\(=\sqrt{\left(2cos^2x+1\right)^2}+\sqrt{\left(2sin^2x+1\right)^2}\)
\(=2cos^2x+1+2sin^2x+1\)
\(=2\left(sin^2x+cos^2x\right)+2=4\)
1 tháng 7 2021
a)\(-1\le sinx\le1\)
\(\Leftrightarrow1\ge-sinx\ge-1\)
\(\Leftrightarrow4\ge3-sinx\ge2\) \(\Leftrightarrow16\ge\left(3-sinx\right)^2\ge4\)\(\Leftrightarrow17\ge\left(3-sinx\right)^2+1\ge5\)
\(\Leftrightarrow17\ge y\ge5\)
\(y_{min}=5\Leftrightarrow sinx=1\)\(\Leftrightarrow\)\(x=\dfrac{\pi}{2}+k2\pi\)\(\left(k\in Z\right)\)
\(y_{max}=17\Leftrightarrow\)\(sinx=-1\Leftrightarrow x=-\dfrac{\pi}{2}+k2\pi\)\(\left(k\in Z\right)\)
b)\(y=\left(sin^2x+cos^2x\right)^2-2.sinx^2cos^2x\)\(=1-\dfrac{1}{2}.sin^22x\)
Có \(0\le sin^22x\le1\)\(\Leftrightarrow0\ge-\dfrac{1}{2}.sin^22x\ge-\dfrac{1}{2}\)
\(\Leftrightarrow1\ge1-\dfrac{1}{2}.sin^22x\ge\dfrac{1}{2}\)\(\Leftrightarrow1\ge y\ge\dfrac{1}{2}\)
\(y_{min}=\dfrac{1}{2}\Leftrightarrow sin^22x=1\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}sin2x=-1\\sin2x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)
\(y_{max}=1\Leftrightarrow sin2x=0\Leftrightarrow x=\dfrac{k\pi}{2}\)\(\left(k\in Z\right)\)
c)\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=1-3sin^2x.cos^2x=1-\dfrac{3}{4}.sin^22x\)
Có \(0\le sin^22x\le1\)\(\Leftrightarrow0\ge-\dfrac{3}{4}.sin^22x\ge-\dfrac{3}{4}\)
\(\Leftrightarrow1\ge1-\dfrac{3}{4}.sin^22x\ge\dfrac{1}{4}\)\(\Leftrightarrow1\ge y\ge\dfrac{1}{4}\)
\(y_{min}=\dfrac{1}{4}\Leftrightarrow sin^22x=1\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(y_{max}=1\Leftrightarrow sin2x=0\Leftrightarrow x=\dfrac{k\pi}{2}\)\(\left(k\in Z\right)\)
Vậy...
HP
1 tháng 7 2021
a, Đặt \(t=sinx\left(t\in\left[-1;1\right]\right)\)
\(y=f\left(t\right)=\left(3-t\right)^2+1=t^2-6t+10\)
\(\Rightarrow min=min\left\{f\left(-1\right);f\left(1\right)\right\}=f\left(1\right)=5\)
\(\Rightarrow max=max\left\{f\left(-1\right);f\left(1\right)\right\}=f\left(-1\right)=17\)
b, \(y=sin^4x+cos^4x=1-2sin^2x.cos^2x=1-\dfrac{1}{2}sin^22x\)
Đặt \(t=sin2x\left(t\in\left[-1;1\right]\right)\)
\(y=f\left(t\right)=1-\dfrac{1}{2}t^2\)
\(\Rightarrow min=min\left\{f\left(-1\right);f\left(0\right);f\left(1\right)\right\}=\dfrac{1}{2}\)
\(\Rightarrow max=max\left\{f\left(-1\right);f\left(0\right);f\left(1\right)\right\}=1\)
c, \(y=sin^6x+cos^6x\)
\(=sin^4x+cos^4x-sin^2x.cos^2x\)
\(=1-3sin^2x.cos^2x\)
\(=1-\dfrac{3}{4}sin^22x\)
Đặt \(t=sin2x\left(t\in\left[-1;1\right]\right)\)
\(y=f\left(t\right)=1-\dfrac{3}{4}t^2\)
\(\Rightarrow min=min\left\{f\left(-1\right);f\left(0\right);f\left(1\right)\right\}=\dfrac{1}{4}\)
\(\Rightarrow max=max\left\{f\left(-1\right);f\left(0\right);f\left(1\right)\right\}=1\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 6 2020
\(B=cos^2x.cot^2x+cos^2x-cot^2x+2\left(sin^2x+cos^2x\right)\)
\(=cos^2x\left(cot^2x+1\right)-cot^2x+2\)
\(=\frac{cos^2x}{sin^2x}-cot^2x+1=cot^2x-cot^2x+1=1\)
\(M=cos^4x-sin^4x+cos^4x+sin^2x.cos^2x+3sin^2x\)
\(=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+cos^2x\left(cos^2x+sin^2x\right)+3sin^2x\)
\(=cos^2x-sin^2x+cos^2x+3sin^2x\)
\(=2\left(sin^2x+cos^2x\right)=2\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 8 2020
6.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)
\(\Leftrightarrow-3sin^22x+sin2x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 8 2020
5.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)
\(\Leftrightarrow sin^22x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 8 2020
4.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=cos2x\)
\(\Leftrightarrow1-\frac{1}{2}sin^22x=cos2x\)
\(\Leftrightarrow1+1-sin^22x=2cos2x\)
\(\Leftrightarrow1+cos^22x=2cos2x\)
\(\Leftrightarrow\left(cos2x-1\right)^2=0\)
\(\Leftrightarrow cos2x=1\)
\(\Leftrightarrow2x=k2\pi\)
\(\Rightarrow x=k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 8 2020
3.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\frac{1}{2}\)
\(\Leftrightarrow1-\frac{1}{2}\left(2sinx.cosx\right)^2=\frac{1}{2}\)
\(\Leftrightarrow1-sin^22x=0\)
\(\Leftrightarrow cos^22x=0\)
\(\Leftrightarrow cos2x=0\)
\(\Leftrightarrow2x=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
TL
1
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 10 2020
1.
\(\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+5cosx+3=0\)
\(\Leftrightarrow2cos^2x-1+5cosx+3=0\)
\(\Leftrightarrow2cos^2x+5cosx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\cosx=-2\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
\(3\left(1-sin^2x\right)+\left(1-sin^2x\right)sinx=8\left(1+sinx\right)\)
\(\Leftrightarrow\left(1+sinx\right)\left(3-3sinx\right)+\left(1+sinx\right)\left(sinx-sin^2x\right)=8\left(1+sinx\right)\)
\(\Leftrightarrow\left(1+sinx\right)\left(3-3sinx+sinx-sin^2x-8\right)=0\)
\(\Leftrightarrow\left(1+sinx\right)\left(-sin^2x-2sinx-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\-sin^2x-2sinx-5=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
\(3sin^4x-\left(1-sin^2x\right)^2=\frac{1}{2}\Leftrightarrow3sin^4x-\left(sin^4x-2sin^2x+1\right)=\frac{1}{2}\)
\(\Leftrightarrow2sin^4x+2sin^2x-\frac{3}{2}=0\) \(\Rightarrow\left[{}\begin{matrix}sin^2x=\frac{1}{2}\\sin^2x=-\frac{3}{2}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow cos^2x=1-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow B=\left(\frac{1}{2}\right)^2+3\left(\frac{1}{2}\right)^2=1\)
\(4sin^4x+3\left(1-sin^2x\right)^2=\frac{7}{4}\Leftrightarrow4sin^4x+3\left(sin^4x-2sin^2x+1\right)=\frac{7}{4}\)
\(\Leftrightarrow7sin^4x-6sin^2x+\frac{5}{4}=0\Rightarrow\left[{}\begin{matrix}sin^2x=\frac{1}{2}\Rightarrow cos^2x=\frac{1}{2}\\sin^2x=\frac{5}{14}\Rightarrow cos^2x=\frac{9}{14}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}C=3\left(\frac{1}{2}\right)^2+4\left(\frac{1}{2}\right)^2=\frac{7}{4}\\C=3\left(\frac{5}{14}\right)^2+4\left(\frac{9}{14}\right)^2=\frac{57}{28}\end{matrix}\right.\)