tìm gtln(nếu có)
y=\(\dfrac{3sin^2x\left(1-4sin^2x\right)}{cos^4x}\)
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NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 7 2021
1a.
Đặt \(5x+6=u\)
\(cos2u+4\sqrt{2}sinu-4=0\)
\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)
\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+6=\dfrac{\pi}{4}+k2\pi\\5x+6=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 7 2021
1b.
Đặt \(2x+1=u\)
\(cos2u+3sinu=2\)
\(\Leftrightarrow1-2sin^2u+3sinu=2\)
\(\Leftrightarrow2sin^2u-3sinu+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinu=1\\sinu=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x+1\right)=1\\sin\left(2x+1\right)=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{\pi}{2}+k2\pi\\2x+1=\dfrac{\pi}{6}+k2\pi\\2x+1=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)
ND
Nguyễn Đức Trí
VIP
21 tháng 9 2023
a) \(sin^2x+\left(1-\sqrt[]{3}\right)sinxcosx-\sqrt[]{3}cos^2x=0\)
\(\Leftrightarrow tan^2x+\left(1-\sqrt[]{3}\right)tanx-\sqrt[]{3}=0\left(cosx\ne0\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\sqrt[]{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=tan\dfrac{3\pi}{4}\\tanx=tan\dfrac{\pi}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=tan\dfrac{3\pi}{4}+k\pi\\x=\dfrac{\pi}{3}+k\pi\end{matrix}\right.\left(k\in Z\right)\)
5 tháng 7 2021
1,\(A=3\left(sin^4x+cos^4x\right)-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)
\(=3\left(sin^4x+cos^4x\right)-2\left(sin^4x-sin^2x.cos^4x+cos^4x\right)\)
\(=sin^4x+2sin^2x.cos^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\)
Vậy...
2,\(B=cos^6x+2sin^4x\left(1-sin^2x\right)+3\left(1-cos^2x\right)cos^4x+sin^4x\)
\(=-2cos^6x+3sin^4x-2sin^6x+3cos^4x\)
\(=-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)
\(=-2\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)\(=cos^4x+sin^4x+2sin^2x.cos^2x=1\)
Vậy...
3,\(C=\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}\right)\right]+\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)
\(=cos\left(-\dfrac{7\pi}{12}\right)+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}+\pi\right)\right]\)
\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)-cos\left(2x-\dfrac{\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}\)
Vậy...
4, \(D=cos^2x+\left(-\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx\right)^2+\left(-\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right)^2\)
\(=cos^2x+\dfrac{1}{4}cos^2x+\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x+\dfrac{1}{4}cos^2x-\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x\)
\(=\dfrac{3}{2}\left(cos^2x+sin^2x\right)=\dfrac{3}{2}\)
Vậy...
5, Xem lại đề
6,\(F=-cosx+cosx-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\pi+\dfrac{\pi}{2}-x\right)\)
\(=tan\left(\pi-\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=tan\left(\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=cotx.tanx=1\)
Vậy...
28 tháng 5 2021
a/ \(\left(2sinx-cosx\right)\left(1+cosx\right)=sin^2x\)
\(\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=\dfrac{1-cos2x}{2}\)
\(\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=\dfrac{1-2cos^2x+1}{2}=\dfrac{2-2cos^2x}{2}=1-cos^2x\)
\(\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=\left(1-cosx\right)\left(1+cosx\right)\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)-\left(1-cosx\right)\left(1+cosx\right)=0\)\(\Leftrightarrow\left(1+cosx\right)\left(2sinx-cosx-1+cosx\right)=0\Leftrightarrow\left(1+cosx\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1+cosx=0\\2sinx-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\\sinx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=180^o\\x=30^o\end{matrix}\right.\)
28 tháng 5 2021
a) Đáp án: \(\left[{}\begin{matrix}cosx=-1\\sinx=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)(\(k\in Z\))
Vậy...
b) \(3sin^2x+7cos2x-3=0\)
\(\Leftrightarrow3sin^2x+7\left(1-2sin^2x\right)-3=0\)
\(\Leftrightarrow11.sin^2x=4\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{2\sqrt{11}}{11}\\sinx=\dfrac{-2\sqrt{11}}{11}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=arc.sin\dfrac{2\sqrt{11}}{11}+k2\pi\\x=\pi-arc.sin\dfrac{2\sqrt{11}}{11}+k2\pi\\x=arc.sin\dfrac{-2\sqrt{11}}{11}+k2\pi\\x=\pi-arc.sin\dfrac{-2\sqrt{11}}{11}+k2\pi\end{matrix}\right.\) (\(k\in Z\)) (Dị quá,câu này e ko biết đ/a đúng hay sai đâu)
Vậy...
c)\(\dfrac{4.sin^2x+6.sin^2x-9-3.cos2x}{cosx}=0\) (đk: \(x\ne\dfrac{\pi}{2}+k\pi\),\(k\in Z\))
\(\Rightarrow10sin^2x-9-3\left(1-2.sin^2x\right)=0\)
\(\Leftrightarrow sin^2x=\dfrac{3}{4}\)\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{\sqrt{3}}{2}\\sinx=-\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\\x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.\)(\(k\in Z\)) (Thỏa mãn đk)
Vậy...
HP
17 tháng 9 2021
a, \(y=3-4sin^2x.cos^2x=3-sin^22x\)
Đặt \(sin2x=t\left(t\in\left[-1;1\right]\right)\).
\(\Rightarrow y=f\left(t\right)=3-t^2\)
\(\Rightarrow y_{min}=minf\left(t\right)=2\)
\(y_{max}=maxf\left(t\right)=3\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 3 2019
Giả sử các biểu thức đều có nghĩa
\(A=2\left(\left(sin^2x\right)^3+\left(cos^2x\right)^3\right)-3\left(sin^4x+cos^4x+2sin^2xcos^2x-2sin^2xcos^2x\right)\)
\(A=2\left(sin^2x+cos^2x\right)\left(\left(sin^2x+cos^2x\right)^2-3sin^2xcos^2x\right)-3\left(\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\right)\)
\(A=2\left(1-3sin^2xcos^2x\right)-3\left(1-2sin^2xcos^2x\right)\)
\(A=2-6sin^2xcos^2x-3+6sin^2xcos^2x=-1\)
b/ \(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{tanx-1}=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{\dfrac{1}{cotx}-1}\)
\(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2cotx}{1-cotx}=\dfrac{1+cotx-2cotx}{1-cotx}=\dfrac{1-cotx}{1-cotx}=1\)
c/ \(C=cos^4x-sin^4x+cos^4x+sin^2xcos^2x+3sin^2x\)
\(C=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+cos^2x\left(cos^2x+sin^2x\right)+3sin^2x\)
\(C=cos^2x-sin^2x+cos^2x+3sin^2x\)
\(C=2cos^2x+2sin^2x=2\left(cos^2x+sin^2x\right)=2\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 7 2021
a.
\(y=\dfrac{3}{2}sin2x-2\left(cos^2x-sin^2x\right)+5=\dfrac{3}{2}sin2x-2cos2x+5\)
\(=\dfrac{5}{2}\left(\dfrac{3}{5}sin2x-\dfrac{4}{5}cos2x\right)+5=\dfrac{5}{2}sin\left(2x-a\right)+5\) (với \(cosa=\dfrac{3}{5}\))
\(\Rightarrow-\dfrac{5}{2}+5\le y\le\dfrac{5}{2}+5\)
b.
\(\Leftrightarrow y.sinx-2y.cosx+4y=3sinx-cosx+1\)
\(\Leftrightarrow\left(y-3\right)sinx+\left(1-2y\right)cosx=1-4y\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(\left(y-3\right)^2+\left(1-2y\right)^2\ge\left(1-4y\right)^2\)
\(\Leftrightarrow11y^2+2y-9\le0\)
\(\Leftrightarrow-1\le y\le\dfrac{9}{11}\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 7 2021
c.
Do \(x^2+y^2=1\Rightarrow\) đặt \(\left\{{}\begin{matrix}x=sina\\y=cosa\end{matrix}\right.\)
\(\Rightarrow y=\dfrac{2\left(sin^2a+6sina.cosa\right)}{1+2sina.cosa+cos^2a}=\dfrac{1-cos2a+6sin2a}{1+sin2a+\dfrac{1+cos2a}{2}}=\dfrac{2-2cos2a+12sin2a}{3+2sin2a+cos2a}\)
\(\Leftrightarrow3y+2y.sin2a+y.cos2a=2-2cos2a+12sin2a\)
\(\Leftrightarrow\left(2y-12\right)sin2a+\left(y+2\right)cos2a=2-3y\)
Theo điều kiện có nghiệm của pt bậc nhất theo sin2a, cos2a:
\(\left(2y-12\right)^2+\left(y+2\right)^2\ge\left(2-3y\right)^2\)
\(\Leftrightarrow y^2+8y-36\le0\)
\(\Rightarrow-4-2\sqrt{13}\le y\le-4+2\sqrt{13}\)
\(y=\dfrac{3sin^2x-12sin^4x}{cos^4x}=3tan^2x.\dfrac{1}{cos^2x}-12tan^4x=3tan^2x\left(1+tan^2x\right)-12tan^4x\)
\(=-9tan^4x+3tan^2x=-9\left(tan^2x-\dfrac{1}{6}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)