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e/
\(\Leftrightarrow\left(sin^2x+4sinx.cosx+3cos^2x\right)-\left(sinx+3cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sinx+3cosx\right)-\left(sinx+3cosx\right)=0\)
\(\Leftrightarrow\left(sinx+3cosx\right)\left(sinx+cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+3cosx=0\\sinx+cosx-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-3cosx\\\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-3\\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(-3\right)+k\pi\\x=k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
d/
\(\Leftrightarrow2sinx+2sinx.cos2x-\left(1-sin2x\right)-2cosx=0\)
\(\Leftrightarrow2\left(sinx-cosx\right)+2sinx\left(cos^2x-sin^2x\right)-\left(sinx-cosx\right)^2=0\)
\(\Leftrightarrow2\left(sinx-cosx\right)-2sinx\left(sinx-cosx\right)\left(sinx+cosx\right)-\left(sinx-cosx\right)^2=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(2-2sin^2x-2sinx.cosx-sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left[2cos^2x-2sinx.cosx-sinx+cosx\right]=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left[2cosx\left(cosx-sinx\right)+cosx-sinx\right]=0\)
\(\Leftrightarrow-\left(sinx-cosx\right)^2\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\2cosx+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
e/
\(y=5sinx+6cosx-7\)
\(=\sqrt{61}\left(\frac{5}{\sqrt{61}}sinx+\frac{6}{\sqrt{61}}cosx\right)-7\)
\(=\sqrt{61}\left(sinx.cosa+cosx.sina\right)-7\) (với \(a\in\left(0;\pi\right)\) sao cho \(cosa=\frac{5}{\sqrt{61}}\))
\(=\sqrt{61}.sin\left(x+a\right)-7\)
Do \(-1\le sin\left(x+a\right)\le1\Rightarrow7-\sqrt{61}\le y\le7+\sqrt{61}\)
\(y_{min}=7-\sqrt{61}\) khi \(sin\left(x+a\right)=-1\)
\(y_{max}=7+\sqrt{61}\) khi \(sin\left(x+a\right)=1\)
f/
\(y=2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)+3\)
\(=2sin\left(x+\frac{\pi}{3}\right)+3\)
\(\Rightarrow1\le y\le5\)
\(y_{min}=1\) khi \(sin\left(x+\frac{\pi}{3}\right)=-1\)
\(y_{max}=5\) khi \(x+\frac{\pi}{3}=1\)
c/
\(y=2\left(1-cos2x\right)+sin2x+cos2x\)
\(=sin2x-cos2x+2=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)+2\)
Do \(-1\le sin\left(2x-\frac{\pi}{4}\right)\le1\)
\(\Rightarrow2-\sqrt{2}\le y\le2+\sqrt{2}\)
\(y_{min}=2-\sqrt{2}\) khi \(sin\left(2x-\frac{\pi}{4}\right)=-1\)
\(y_{max}=2+\sqrt{2}\) khi \(sin\left(2x+\frac{\pi}{4}\right)=1\)
d/
\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)\)
\(=1-3sin^2x.cos^2x\)
\(=1-\frac{3}{4}sin^22x\)
Mà \(0\le sin^22x\le1\Rightarrow\frac{1}{4}\le y\le1\)
\(y_{min}=\frac{1}{4}\) khi \(sin^22x=1\)
\(y_{max}=1\) khi \(sin2x=0\)
1.
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-\frac{\sqrt{3}}{2}\\cos4x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow x=...\)
(Cứ bấm máy giải pt bậc 2 như bt, nó cho 2 nghiệm rất xấu, bạn lưu 2 nghiệm vào 2 biến A; B rồi thoát ra ngoài MODE-1, tính \(\sqrt{A^2}\) và \(\sqrt{B^2}\) sẽ ra dạng căn đẹp của 2 nghiệm, lưu ý dấu so với nghiệm ban đầu)
2.
\(\Leftrightarrow cos4x+1+sin\left(2x-\frac{\pi}{2}\right)=cos2x\)
\(\Leftrightarrow2cos^22x-cos2x=cos2x\)
\(\Leftrightarrow cos^22x-cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\end{matrix}\right.\)
3.
\(\Leftrightarrow\frac{1}{2}sin\left(x+\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}cos\left[\frac{\pi}{2}-\left(\frac{\pi}{6}-x\right)\right]=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}sin\left(x+\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}cos\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x+\frac{2\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow...\)
4.
\(\Leftrightarrow2cos4x.cos\left(\frac{\pi}{3}\right)+2sin4x.sin\left(\frac{\pi}{3}\right)+4cos2x=-1\)
\(\Leftrightarrow cos4x+\sqrt{3}sin4x+4cos2x+1=0\)
\(\Leftrightarrow2cos^22x+2\sqrt{3}sin2x.cos2x+4cos2x=0\)
\(\Leftrightarrow2cos2x\left(cos2x+\sqrt{3}sin2x+2\right)=0\)
\(\Leftrightarrow cos2x\left(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x+1\right)=0\)
\(\Leftrightarrow cos2x\left[sin\left(2x+\frac{\pi}{6}\right)+1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(2x+\frac{\pi}{6}\right)=-1\end{matrix}\right.\)
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) làm tương tự 2 bài mk đã giải nha.
b) \(y=2\cos^2x-2\sqrt{3}\sin x\cos x+1\)
\(=1-\left(\cos2x+\sqrt{3}\sin2x\right)\)
Lại có \(-2\le\cos2x+\sqrt{3}\sin2x\le2\) \(\Rightarrow-1\le y\le3\)
c) Vì \(\left\{{}\begin{matrix}0\le\sqrt[4]{\sin x}\le1\\0\le\sqrt{\cos x}\le1\end{matrix}\right.\)
Do đó \(-1\le y\le1\)
3.
a.
\(\Leftrightarrow\left(cos3x-cosx\right)+\left(cos2x-1\right)=0\)
\(\Leftrightarrow-2sin2x.sinx+1-2sin^2x-1=0\)
\(\Leftrightarrow sin2x.sinx+sin^2x=0\)
\(\Leftrightarrow2sin^2x.cosx+sin^2x=0\)
\(\Leftrightarrow sin^2x\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
e/
Đề câu này chắc chắn đúng chứ bạn?
f/
\(sin^4x+cos^4x=\frac{3}{4}\)
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\frac{3}{4}\)
\(\Leftrightarrow1-\frac{1}{2}\left(2sinx.cosx\right)^2=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{1}{2}sin^22x=0\)
\(\Leftrightarrow1-2sin^22x=0\)
\(\Leftrightarrow cos4x=0\)
\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)
c/
\(y=sin\left(4x-\frac{\pi}{3}\right)+sin\left(\frac{\pi}{3}\right)+5\)
\(=sin\left(4x-\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}+5\)
Do \(-1\le sin\left(4x-\frac{\pi}{3}\right)\le1\)
\(\Rightarrow4+\frac{\sqrt{3}}{2}\le y\le6+\frac{\sqrt{3}}{2}\)
d/
\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+3sin2x+5\)
\(y=6-3sin^2x.cos^2x+3sin2x\)
\(y=-\frac{3}{4}sin^22x+3sin2x+6\)
\(y=\frac{3}{4}\left(sin2x+1\right)\left(5-sin2x\right)+\frac{9}{4}\ge\frac{9}{4}\)
\(y_{min}=\frac{9}{4}\) khi \(sin2x=-1\)
\(y=\frac{3}{4}\left(sin2x-1\right)\left(3-sin2x\right)+\frac{33}{4}\le\frac{33}{4}\)
\(y_{max}=\frac{33}{4}\) khi \(sin2x=1\)
a.
\(-1\le sinx\le1\Rightarrow-7\le y\le-3\)
\(y_{min}=-7\) khi \(sinx=-1\)
\(y_{max}=-3\) khi \(sinx=1\)
b.
\(-1\le cos\left(x+\frac{\pi}{3}\right)\le1\Rightarrow1\le y\le5\)
\(y_{min}=1\) khi \(cos\left(x+\frac{\pi}{3}\right)=-1\)
\(y_{max}=5\) khi \(cos\left(x+\frac{\pi}{3}\right)=1\)
c.
\(0\le1-cosx\le2\Rightarrow-5\le y\le3\sqrt{2}-5\)
\(y_{min}=-5\) khi \(cosx=1\)
\(y_{max}=3\sqrt{2}-5\) khi \(cosx=-1\)
d.
ĐKXĐ: \(0\le sinx\Rightarrow0\le sinx\le1\Rightarrow1\le y\le3\)
\(y_{min}=1\) khi \(sinx=0\)
\(y_{max}=3\) khi \(sinx=1\)