a/
\(\Leftrightarrow sinx+cosx-4sinx.cosx-1=0\)
Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(\Rightarrow t^2=1+2sinx.cosx\Rightarrow sinx.cosx=\frac{t^2-1}{2}\)
Pt trở thành:
\(t-2\left(t^2-1\right)-1=0\)
\(\Leftrightarrow-2t^2+t+1=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\\\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\\sin\left(x+\frac{\pi}{4}\right)=-\frac{1}{2\sqrt{2}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x+\frac{\pi}{4}=arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\\x+\frac{\pi}{4}=\pi-arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\) \(\Rightarrow x=...\)
b/
Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=t\Rightarrow sinx.cosx=\frac{t^2-1}{2}\)
Pt trở thành:
\(t+\frac{3}{2}\left(t^2-1\right)-1=0\)
\(\Leftrightarrow3t^2+2t-5=0\)
\(\Rightarrow\left[{}\begin{matrix}t=-1\\t=\frac{5}{3}>\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow sinx+cosx-4sinx.cosx=\frac{1}{2}\)
Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=t\) với \(\left|t\right|\le\sqrt{2}\)
\(sinx.cosx=\frac{t^2-1}{2}\)
Pt trở thành:
\(t-2\left(t^2-1\right)=\frac{1}{2}\)
\(\Leftrightarrow-4t^2+2t+3=0\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{1+\sqrt{13}}{4}\\t=\frac{1-\sqrt{13}}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=\frac{1+\sqrt{13}}{4\sqrt{2}}\\sin\left(x+\frac{\pi}{4}\right)=\frac{1-\sqrt{13}}{4\sqrt{2}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=arcsin\left(\frac{1+\sqrt{13}}{4\sqrt{2}}\right)+k2\pi\\x+\frac{\pi}{4}=\pi-arcsin\left(\frac{1+\sqrt{13}}{4\sqrt{2}}\right)+k2\pi\\x+\frac{\pi}{4}=arcsin\left(\frac{1-\sqrt{13}}{4\sqrt{2}}\right)+k2\pi\\x+\frac{\pi}{4}=\pi-arcsin\left(\frac{1-\sqrt{13}}{4\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)
d/
Đặt \(sinx-cosx=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\) \(\Rightarrow\left|t\right|\le\sqrt{2}\)
\(t^2=1-2sinx.cosx\Rightarrow sinx.cosx=\frac{1-t^2}{2}\)
Pt trở thành:
\(6t-1=\frac{1-t^2}{2}\)
\(\Leftrightarrow t^2+12t-3=0\)
\(\Rightarrow\left[{}\begin{matrix}t=\sqrt{39}-6\\t=-\sqrt{39}-6< -\sqrt{2}\left(l\right)\end{matrix}\right.\) (ủa giáo viên ra đề ngẫu nhiên à?)
\(\Rightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{39}-6}{\sqrt{2}}\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=arcsin\left(\frac{\sqrt{39}-6}{\sqrt{2}}\right)+k2\pi\\x-\frac{\pi}{4}=\pi-arcsin\left(\frac{\sqrt{39}-6}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)
giúp mk vs, (sr mk k đăng đc)
bài 1: Giải các PT:
a) \(cos^2x-\sqrt{3}sin2x=sin^2x+1\)
b) \(2cos^2x+5sinx.cosx+6sin^2x-1=0\)
c) \(5cos^2x+\sqrt{3}sinx.cosx-2=0\)
d) \(cosx-\sqrt{3}sinx=\frac{1}{cosx}\)
e) \(2cos^2x-\left(sinx+cosx\right)^2-4sin^2x=0\)
bài 2: Giải các PT:
a) \(tanx+cotx=2\left(sin2x+cos2x\right)\)
b) \(2\sqrt{2}\left(sinx+cosx\right).cosx=3+2cos^2x\)
c) \(\left(cosx-2sinx\right)^2=\frac{1}{2}+cos2x\)
d) \(4\frac{sin^2x+3\sqrt{3}sin2x-cos^2x}{4+cos^2x}=1\)
e) \(6sin^2x-\left|sin2x\right|-4cos2x=8\)
Bạn đăng rồi đấy, có điều nó ko thông báo thôi
