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a.
\(\Leftrightarrow\left[{}\begin{matrix}3x=90^0-x+k360^0\\3x=90^0+x+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{45^0}{2}+k90^0\\x=45^0+k180^0\end{matrix}\right.\)
b.
\(\Leftrightarrow cos\left(3x+45^0\right)=cos\left(x-180^0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+45^0=x-180^0+k360^0\\3x+45^0=180^0-x+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{225^0}{2}+k180^0\\x=\frac{135^0}{4}+k90^0\end{matrix}\right.\)
c.
\(\Leftrightarrow sin\left(2x+\frac{\pi}{3}\right)=sin\left(-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=-x+k2\pi\\2x+\frac{\pi}{3}=\pi+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{9}+\frac{k2\pi}{3}\\x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
d.
\(\Leftrightarrow sin\left(x-\frac{2\pi}{3}\right)=cos2x\)
\(\Leftrightarrow sin\left(x-\frac{2\pi}{3}\right)=sin\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{2\pi}{3}=\frac{\pi}{2}-x+k2\pi\\x-\frac{2\pi}{3}=2x+\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{12}+k\pi\\x=-\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
e.
\(\Leftrightarrow cos\left(2x-\frac{\pi}{4}\right)=sin\left(2x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(2x-\frac{\pi}{4}\right)=cos\left(\frac{\pi}{6}-2x\right)\)
\(\Leftrightarrow2x-\frac{\pi}{4}=\frac{\pi}{6}-2x+k2\pi\)
\(\Leftrightarrow x=\frac{5\pi}{48}+\frac{k\pi}{2}\)
6.
\(\Leftrightarrow\frac{1}{2}cos6x+\frac{1}{2}cos4x=\frac{1}{2}cos6x+\frac{1}{2}cos2x+\frac{3}{2}+\frac{3}{2}cos2x+1\)
\(\Leftrightarrow cos4x=4cos2x+5\)
\(\Leftrightarrow2cos^22x-1=4cos2x+5\)
\(\Leftrightarrow cos^22x-2cos2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=3>1\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
7.
Thay lần lượt 4 đáp án ta thấy chỉ có đáp án C thỏa mãn
8.
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=\left\{\frac{\pi}{6};\frac{\pi}{2}\right\}\)
9.
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}-1\le t\le1\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Rightarrow mt+\frac{t^2-1}{2}+1=0\)
\(\Leftrightarrow t^2+2mt+1=0\)
Pt đã cho có đúng 1 nghiệm thuộc \(\left[-1;1\right]\) khi và chỉ khi: \(\left[{}\begin{matrix}m\ge1\\m\le-1\end{matrix}\right.\)
10.
\(\frac{\sqrt{3}}{2}cos5x-\frac{1}{2}sin5x=cos3x\)
\(\Leftrightarrow cos\left(5x-\frac{\pi}{6}\right)=cos3x\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\frac{\pi}{6}=3x+k2\pi\\5x-\frac{\pi}{6}=-3x+k2\pi\end{matrix}\right.\)
a.
\(cos\left(3x-\frac{\pi}{6}\right)=sin\left(2x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(3x-\frac{\pi}{6}\right)=cos\left(\frac{\pi}{6}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=\frac{\pi}{6}-2x+k2\pi\\3x-\frac{\pi}{6}=2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\cos3x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne\frac{1}{2}\end{matrix}\right.\)
\(tan3x-tanx=0\)
\(\Leftrightarrow\frac{sin3x}{cos3x}-\frac{sinx}{cosx}=0\)
\(\Leftrightarrow sin3x.cosx-cos3x.sinx=0\)
\(\Leftrightarrow sin2x=0\)
\(\Leftrightarrow2sinx.cosx=0\)
\(\Leftrightarrow sinx=0\Leftrightarrow x=k\pi\)
c.
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{2\pi}{5}\right)=\frac{1}{2}-\frac{1}{2}cos\left(4x+\frac{8\pi}{5}\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=-cos\left(4x+\frac{3\pi}{5}+\pi\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=cos\left(4x+\frac{3\pi}{5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{3\pi}{5}=2x-\frac{2\pi}{5}+k2\pi\\4x+\frac{3\pi}{5}=\frac{2\pi}{5}-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
d.
\(\Leftrightarrow cos^2\left(2x-1\right)=0\)
\(\Leftrightarrow cos\left(2x-1\right)=0\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{1}{2}+\frac{k\pi}{2}\)
ngại viết quá hihi, mà hơi ngáo tí cái dạng này lm rồi mà cứ quên
bài trước mk bình luận bạn đọc chưa nhỉ
a/
\(y=\frac{1}{sinx}+\frac{1}{cosx}\ge\frac{4}{sinx+cosx}=\frac{4}{\sqrt{2}sin\left(x+\frac{\pi}{4}\right)}\ge\frac{4}{\sqrt{2}}=2\sqrt{2}\)
\(y_{min}=2\sqrt{2}\) khi \(\left\{{}\begin{matrix}sinx=cosx\\sin\left(x+\frac{\pi}{4}\right)=1\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{4}\)
\(y_{max}\) không tồn tại (y dần tới dương vô cùng khi x gần tới 0 hoặc \(\frac{\pi}{2}\))
b/
\(y=\frac{1}{1-cosx}+\frac{1}{1+cosx}=\frac{1+cosx+1-cosx}{1-cos^2x}=\frac{2}{sin^2x}\)
Hàm số ko tồn tại cả min lẫn max ( \(0< y< \infty\))
c/
Do \(tan^2x\) ko tồn tại max (tiến tới vô cực) trên khoảng đã cho nên hàm ko tồn tại max
\(y=2+\frac{sin^4x+cos^4x}{\left(sinx.cosx\right)^2}+\frac{1}{sin^4x+cos^4x}\ge2+2\sqrt{\frac{sin^4x+cos^4x}{\frac{1}{4}sin^22x.\left(sin^4x+cos^4x\right)}}\)
\(y\ge2+\frac{4}{sin2x}\ge2+\frac{4}{1}=6\)
\(y_{min}=6\) khi \(\left\{{}\begin{matrix}sin2x=1\\sin^4x+cos^4x=sinx.cosx\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{4}\)
c. ĐKXĐ: ...
\(\Leftrightarrow\frac{\pi}{3}sin\pi x=\frac{\pi}{6}+k\pi\)
\(\Leftrightarrow sin\pi x=\frac{1}{2}+3k\)
\(-1\le\frac{1}{2}+3k\le1\Rightarrow k=0\)
\(\Rightarrow sin\pi x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\pi x=\frac{\pi}{6}+k2\pi\\\pi x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{6}+2k\\x=\frac{5}{6}+2k\end{matrix}\right.\)
a/
\(\Leftrightarrow\frac{\pi}{6}cosx+\frac{\pi}{3}=k\pi\)
\(\Leftrightarrow cosx=-2+6k\)
Do \(-1\le cosx\le1\Rightarrow-1\le-2+6k\le1\)
\(\Rightarrow\frac{1}{6}\le k\le\frac{1}{2}\Rightarrow\) ko tồn tại k thỏa mãn
Vậy pt vô nghiệm
b.
\(\Leftrightarrow\pi cos3x=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow cos3x=\frac{1}{2}+k\)
\(-1\le\frac{1}{2}+k\le1\Rightarrow k=\left\{-1;0\right\}\)
\(\Rightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\cosx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{3}+k2\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
3) 2sin^2 x - 3sinx + 1 = 0
Đặt t = sin x
(*) <=> 2t^2 - 3t + 1 = 0
<=> t = 1 (nhận) or t = 1/2 (nhận)
.Vs t = 1 => sinx = 1
<=> x = π/2 + k2π (k thuộc Z) (nhận)
.Vs t = 1/2 => sinx = 1/2
<=> sinx = sin π/6
<=> x = π/6 + k2π (k thuộc Z) (nhận)
Vậy ...
2) cos^2 x + cosx = 0
Đặt t = cosx
(*) <=> t^2 + t =0 <=> t = 0 (n) or t = -1 (n)
. Vs t = 0 => cosx = 0 <=> x = π/2 + kπ (loại)
.Vs t = -1 => cosx = -1 <=> x = π + k2π (nhận)
Vậy ...
1) (sin3x)/cosx + 1 = 0
ĐK: cosx + 1 ≠ 0 <=> cosx ≠ -1 <=> x ≠ π + k2π
<=> sin3x = 0
<=> 3x = kπ
<=> x = 1/3 kπ (k thuộc Z) (n)
Vậy ...