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\(\Leftrightarrow8sinx.cosx\left(cos^2x-sin^2x\right)-\left(1+cos4x\right)=1\)
\(\Leftrightarrow4sin2x.cos2x-cos4x=2\)
\(\Leftrightarrow2sin4x-cos4x=2\)
\(\Leftrightarrow\frac{2}{\sqrt{5}}sin4x-\frac{1}{\sqrt{5}}cos4x=\frac{2}{\sqrt{5}}\)
\(\Leftrightarrow sin\left(4x+\alpha\right)=\frac{2}{\sqrt{5}}\)
\(\Leftrightarrow...\)
Nghiệm xấu quá, bạn tự giải nốt
\(\Leftrightarrow\left(2cosx-sinx\right)\left(1+sinx\right)=1-sin^2x\)
\(\Leftrightarrow\left(2cosx-sinx\right)\left(1+sinx\right)-\left(1-sinx\right)\left(1+sinx\right)=0\)
\(\Leftrightarrow\left(1+sinx\right)\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\\x=\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
Nghiệm dương nhỏ nhất là \(x=\frac{\pi}{3}\)
Cho e hỏi là vì sao khúc cuối có dấu bằng mà trên đề k có dấu bằng ạ?
Vì mình lấy giá trị nguyên bạn
Chính xác là \(-\frac{1}{4}< k< \frac{2020-\frac{\pi}{2}}{2\pi}\)
\(\Rightarrow-0,25< k< 321,243\) (1)
Nhưng k nguyên nên chỉ cần lấy khoảng ở số nguyên gần nhất, tức là \(0\le k\le321\)
2.
\(\Leftrightarrow1-2cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)+sin\frac{x}{2}sinx-cos\frac{x}{2}sin^2x=0\)
\(\Leftrightarrow-cos\left(\frac{\pi}{2}-x\right)+sinx\frac{x}{2}sinx-cosx\frac{x}{2}sin^2x=0\)
\(\Leftrightarrow-sinx+sin\frac{x}{2}sinx-cos\frac{x}{2}sin^2x=0\)
\(\Leftrightarrow sinx\left(sin\frac{x}{2}-1-cos\frac{x}{2}sinx\right)=0\)
\(\Leftrightarrow sinx\left(sin\frac{x}{2}-1-2cos^2\frac{x}{2}sin\frac{x}{2}\right)=0\)
\(\Leftrightarrow sinx\left(sin\frac{x}{2}-1-2sin\frac{x}{2}\left(1-sin^2\frac{x}{2}\right)\right)=0\)
\(\Leftrightarrow sinx\left(2sin^3\frac{x}{2}-sin\frac{x}{2}-1\right)=0\)
\(\Leftrightarrow sinx\left(sin\frac{x}{2}-1\right)\left(2sin^2\frac{x}{2}+2sin\frac{x}{2}+1\right)=0\)
\(\Leftrightarrow...\)
1.
\(\Leftrightarrow sin2x-4sin\left(x+\frac{\pi}{4}\right)=5\)
Do \(\left\{{}\begin{matrix}sin2x\le1\\-4sin\left(x+\frac{\pi}{4}\right)\le4\end{matrix}\right.\) với mọi x
\(\Rightarrow sin2x-4sin\left(x+\frac{\pi}{4}\right)\le5\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}sin2x=1\\sin\left(x+\frac{\pi}{4}\right)=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow x=-\frac{3\pi}{4}+k2\pi\)
\(\cos5x=-\sin4x\)
<=> \(\cos5x=\cos\left(4x+\frac{\pi}{2}\right)\)
\(\Leftrightarrow\orbr{\begin{cases}5x=4x+\frac{\pi}{2}+k2\pi\\5x=-4x-\frac{\pi}{2}+k2\pi\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{\pi}{2}+k2\pi\\x=-\frac{\pi}{18}+\frac{k2\pi}{9}\end{cases}}\)
Nghiệm âm lớn nhất: \(-\frac{\pi}{18}\)
Nghiệm dương nhỏ nhất: \(\frac{\pi}{2}\)
pt <=> \(\sin\left(5x+\frac{\pi}{3}\right)=\sin\left(2x-\frac{\pi}{3}+\frac{\pi}{2}\right)\)
<=> \(\sin\left(5x+\frac{\pi}{3}\right)=\sin\left(2x+\frac{\pi}{6}\right)\)
<=> \(\orbr{\begin{cases}5x+\frac{\pi}{3}=2x+\frac{\pi}{6}+k2\pi\\5x+\frac{\pi}{3}=\pi-2x-\frac{\pi}{6}+k2\pi\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{\pi}{18}+\frac{k2\pi}{3}\\x=\frac{\pi}{14}+\frac{k2\pi}{7}\end{cases}}\)
Trên \(\left[0,\pi\right]\)có các nghiệm:
\(\frac{11\pi}{18},\frac{\pi}{14},\frac{5\pi}{14},\frac{9\pi}{14},\frac{13\pi}{14}\)
tính tổng:...
1.
\(\Leftrightarrow\left(1-cos6x\right)cos2x+1-cos2x=0\)
\(\Leftrightarrow cos2x-cos2x.cos6x+1-cos2x=0\)
\(\Leftrightarrow\frac{1}{2}\left(cos8x-cos4x\right)-1=0\)
\(\Leftrightarrow2cos^24x-cos4x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-1\\cos4x=\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow4x=\pi+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
3.
Đặt \(\frac{x}{6}=t\Rightarrow\frac{1}{4}+cos^22t=\frac{1}{2}sin^23t\)
\(\Leftrightarrow1+4cos^22t=1-cos6t\)
\(\Leftrightarrow cos6t+4cos^22t=0\)
\(\Leftrightarrow4cos^32t+4cos^22t-3cos2t=0\)
\(\Leftrightarrow cos2t\left(4cos^22t+4cos2t-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2t=0\\cos2t=\frac{1}{2}\\cos2t=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{\pi}{4}+\frac{k\pi}{2}\\t=\pm\frac{\pi}{6}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{3}=\frac{\pi}{4}+\frac{k\pi}{2}\\\frac{x}{3}=\frac{\pi}{6}+k\pi\\\frac{x}{3}=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow x=...\)
d/
ĐKXĐ: ...
Biến đôi biểu thức vế trái trước:
\(1+tanx.tan\frac{x}{2}=1+\frac{sinx.sin\frac{x}{2}}{cosx.cos\frac{x}{2}}=\frac{sinx.sin\frac{x}{2}+cosx.cos\frac{x}{2}}{cosx.cos\frac{x}{2}}=\frac{cos\left(x-\frac{x}{2}\right)}{cosx.cos\frac{x}{2}}=\frac{1}{cosx}\)
Do đó pt tương đương:
\(\sqrt{3}\left(1+tan^2x\right)-tanx-2\sqrt{3}=sinx.\frac{1}{cosx}\)
\(\Leftrightarrow\sqrt{3}tan^2x-2tanx-\sqrt{3}=0\)
\(\Rightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=-\frac{1}{\sqrt{3}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
Sử dụng kết quả biến đổi trên làm câu c sẽ lẹ hơn cách cũ
c/
ĐKXĐ: ...
\(\Leftrightarrow2cos^2x\left(1+tanx.tan\frac{x}{2}\right)=2cos^2x-4\)
\(\Leftrightarrow2cos^2x+2cos^2x.tanx.tan\frac{x}{2}=2cos^2x-4\)
\(\Leftrightarrow cos^2x.tanx.tan\frac{x}{2}=-2\)
\(\Leftrightarrow sinx.cosx.tan\frac{x}{2}=-2\)
\(\Leftrightarrow sinx.cosx.\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=-2\)
\(\Leftrightarrow sinx.cosx.\frac{sin^2\frac{x}{2}}{2sin\frac{x}{2}.cos\frac{x}{2}}=-1\)
\(\Leftrightarrow cosx\left(\frac{1-cosx}{2}\right)=-1\)
\(\Leftrightarrow cos^2x-cosx-2=0\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\pi+k2\pi\)
a/
\(\Leftrightarrow sin2x\left(1+\sqrt{2}sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}sinx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sinx=-\frac{\sqrt{2}}{2}=sin\left(-\frac{\pi}{4}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\x=-\frac{\pi}{4}+k2\pi\\x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=-\frac{\pi}{4}+k2\pi\\x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow2sin2x.cos2x-\frac{1}{2}sin4x+\frac{1}{2}sinx=0\)
\(\Leftrightarrow sin4x-\frac{1}{2}sin4x+\frac{1}{2}sinx=0\)
\(\Leftrightarrow sin4x=-sinx=sin\left(-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}4x=-x+k2\pi\\4x=\pi+x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{5}\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)
e/
\(sin\left(\frac{3\pi}{2}-sinx\right)=1\)
\(\Leftrightarrow\frac{3\pi}{2}-sinx=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow sinx=\pi+k2\pi\)
Mà \(-1\le sinx\le1\Rightarrow-1\le\pi+k2\pi\le1\)
\(\Rightarrow\) Không tồn tại k nguyên thỏa mãn
Pt đã cho vô nghiệm
f/
\(cos^2x-sin^2x+sin4x=0\)
\(\Leftrightarrow cos2x+2sin2x.cos2x=0\)
\(\Leftrightarrow cos2x\left(1+2sin2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin2x=-\frac{1}{2}=sin\left(-\frac{\pi}{6}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x=-\frac{\pi}{6}+k2\pi\\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)