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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/ \(cos^2x-cosx=0\Rightarrow cosx\left(cosx-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=k2\pi\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{2}\)
b/ \(cos2x+sinx=0\Leftrightarrow cos2x=sin\left(-x\right)\)
\(\Leftrightarrow cos2x=cos\left(x+\frac{\pi}{2}\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x=x+\frac{\pi}{2}+k2\pi\\2x=-x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=-\frac{\pi}{6}+\frac{k2\pi}{3}\end{matrix}\right.\)
\(\Rightarrow x=\left\{\frac{\pi}{2};\frac{7\pi}{6};\frac{11\pi}{6}\right\}\)
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.
\(\frac{\sqrt{3}}{2}cos5x-\frac{1}{2}sin5x=cos3x\)
\(\Leftrightarrow cos\left(5x-\frac{\pi}{6}\right)=cos3x\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\frac{\pi}{6}=3x+k2\pi\\5x-\frac{\pi}{6}=-3x+k2\pi\end{matrix}\right.\)
\(a\text{) }sin^3x+cos^3x=sinx+cosx\\ \Leftrightarrow\left(sinx+cosx\right)\left(sin^2x-sinx\cdot cosx+cos^2x\right)=sinx+cosx\\ \Leftrightarrow-\frac{1}{2}sin2x\left(sinx+cosx\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}sinx=-cosx=sin\left(x-\frac{\pi}{2}\right)\\sin2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3\pi}{2}-x+a2\pi\\2x=b\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\frac{3\pi}{4}+a\pi\\x=\frac{b\pi}{2}\end{matrix}\right.\)
\(\text{b) }sin^3x+2sin^2x\cdot cosx-3cos^3x=0\\ \Leftrightarrow\left(sin^3x-cos^3x\right)+2cosx\cdot\left(sin^2x-cos^2x\right)=0\\ \Leftrightarrow\left(sinx-cosx\right)\left(sinx\cdot cosx+1\right)+\left(sinx-cosx\right)\left(2sinx\cdot cosx+2cos^2x\right)=0\\ \Leftrightarrow\left(sinx-cosx\right)\left(3sinx\cdot cosx+1+2cos^2x\right)=0\\ \Leftrightarrow\left(sinx-cosx\right)\left(\frac{3}{2}sin2x+2+cos2x\right)=0\)
Với \(sinx-cosx=0\)
\(\Leftrightarrow sinx=cosx=sin\left(\frac{\pi}{2}-x\right)\\ \Leftrightarrow x=\frac{\pi}{2}-x+a2\pi\\ \Leftrightarrow x=\frac{\pi}{4}+a\pi\)
Với \(\frac{3}{2}sin2x+2+cos2x=0\)
\(\Leftrightarrow sin^22x+\left(\frac{3}{2}sin2x+2\right)^2=1\left(VN\right)\)
\(\text{c) }3cos^4x-4cos^2x\cdot sin^2x-sin^4x=0\)
Nhận thấy sinx=0 không là nghiệm pt.
Chia cả 2 vế cho sin4x ta được
\(pt\Leftrightarrow\frac{3cos^4x}{sin^4x}-\frac{4cos^2x}{sin^2x}-1=0\\ \Leftrightarrow3cot^4x-4cot^2x-1=0\\ \Leftrightarrow cot^2x=\frac{2+\sqrt{7}}{3}\\ \Leftrightarrow cotx=\pm\sqrt{\frac{2+\sqrt{7}}{3}}\\ \Leftrightarrow x=arccot\left(\pm\sqrt{\frac{2+\sqrt{7}}{3}}\right)+k2\pi\)
d) kiểm tra đề.
c/
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\frac{5}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\frac{\pi}{2}+k2\pi\)
d/
\(\Leftrightarrow2cos^2\frac{x}{2}-1+3cos\frac{x}{2}+2=0\)
\(\Leftrightarrow2cos^2\frac{x}{2}+3cos\frac{x}{2}+1=0\)
\(\Rightarrow\left[{}\begin{matrix}cos\frac{x}{2}=-1\\cos\frac{x}{2}=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{x}{2}=\pi+k2\pi\\\frac{x}{2}=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\pi+k4\pi\\x=\pm\frac{4\pi}{3}+k4\pi\end{matrix}\right.\)
a/
\(\Rightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\cosx=-\frac{1}{3}\end{matrix}\right.\) (đặt \(cosx=t\) thành pt bậc 2 rồi bấm máy ra nghiệm thôi)
\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{3}+k2\pi\\x=\pm arccos\left(-\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow6\left(1-sin^2x\right)+5sinx-7=0\)
\(\Leftrightarrow-6sin^2x+5sinx-1=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=\frac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=arcsin\left(\frac{1}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)
a/
\(\left(m+1\right)^2+\left(m-1\right)^2\ge\left(2m+3\right)^2\)
\(\Leftrightarrow2m^2+12m+7\le0\)
\(\Leftrightarrow\frac{-6-\sqrt{22}}{2}\le m\le\frac{-6+\sqrt{22}}{2}\)
b/ \(\Leftrightarrow\left\{{}\begin{matrix}m\ge0\\\left(m-1\right)^2+4m\ge m^4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ge0\\m^4-\left(m+1\right)^2\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ge0\\\left(m^2+m+1\right)\left(m^2-m-1\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow0\le m\le\frac{1+\sqrt{5}}{2}\)
c/ \(\Leftrightarrow\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x+\frac{1}{2}=m\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)+\frac{1}{2}=m\)
Do \(-\frac{1}{2}\le sin\left(2x-\frac{\pi}{3}\right)\le\frac{3}{2}\Rightarrow-\frac{1}{2}\le m\le\frac{3}{2}\)
a/ \(2\left(1-cos^2x\right)+3cos^2x-2=m\)
\(\Leftrightarrow cos^2x=m\)
Do \(0\le cos^2x\le1\) nên pt có nghiệm khi và chỉ khi \(0\le m\le1\)
b/ \(\Leftrightarrow\left\{{}\begin{matrix}cosx=m\\sinx\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}cosx=m\\cosx\ne\pm1\end{matrix}\right.\)
\(\Rightarrow-1< m< 1\)