Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
:v bn ns v là bn bik hết là dạng gì rr mà lm ko đc á :))
1: \(\Leftrightarrow4\cdot\dfrac{1+\cos2x}{2}-6\cdot\dfrac{1-\cos2x}{2}+5\sin2x-4=0\)
\(\Leftrightarrow2+2\cos2x-3+3\cos2x+5\sin2x-4=0\)
\(\Leftrightarrow5\sin2x+5\cos2x=5\)
\(\Leftrightarrow\cos2x+\sin2x=1\)
\(\Leftrightarrow\sqrt{2}\cdot\sin\left(2x+\dfrac{\Pi}{4}\right)=1\)
\(\Leftrightarrow\sin\left(2x+\dfrac{\Pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{\Pi}{4}=\dfrac{\Pi}{4}+k2\Pi\\2x+\dfrac{\Pi}{4}=\dfrac{3\Pi}{4}+k2\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=k\Pi\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)
2: \(\Leftrightarrow\sqrt{3}\cdot\dfrac{1+\cos2x}{2}+\sin2x-\sqrt{3}\cdot\dfrac{1-\cos2x}{2}-1=0\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}\cos2x+\sin2x+\sqrt{3}\cdot\dfrac{\cos2x-1}{2}-1=0\)
\(\Leftrightarrow\sin2x+\dfrac{\sqrt{3}}{2}\cos2x+\dfrac{\sqrt{3}}{2}\cos2x-\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}-2}{2}=0\)
\(\Leftrightarrow\sin2x+\sqrt{3}\cos2x=\dfrac{\sqrt{3}-\sqrt{3}+2}{2}=1\)
\(\Leftrightarrow\sin\left(2x+\dfrac{\Pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{\Pi}{3}=\dfrac{\Pi}{6}+k2\Pi\\2x+\dfrac{\Pi}{3}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{12}\Pi+k\Pi\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)
đúng y như trong đề luôn mà bạn , hay là bạn có tính sai chỗ nào đó rồi không
1. \(4\cos^2x-6\sin^2x+5\sin2x-4=0\)
\(\Leftrightarrow4\cos^2x-6\sin^2x+10\sin x\cos x-4\left(\cos^2x+\sin^2x\right)=0\)
\(\Leftrightarrow10\sin x\cos x-10\sin^2x=0\)
\(\Leftrightarrow10\sin x\left(\cos x-\sin x\right)=0\)
2. \(\sqrt{3}\cos^2x+2\sin x\cos x-\sqrt{3}\sin^2x-1=0\)
\(\Leftrightarrow\left(\sqrt{3}\cos^2x+\sin x\cos x\right)+\left(\sin x\cos x-\sqrt{3}\sin^2x\right)-1=0\)
\(\Leftrightarrow2\cos x\left(\dfrac{\sqrt{3}}{2}\cos x+\dfrac{1}{2}\sin x\right)+2\sin x\left(\dfrac{1}{2}\cos x-\dfrac{\sqrt{3}}{2}\sin x\right)-1=0\)
\(\Leftrightarrow2\cos x.\cos\left(\dfrac{\Pi}{6}-x\right)+2\sin x.\sin\left(\dfrac{\Pi}{6}-x\right)-1=0\)
\(\Leftrightarrow\cos\dfrac{\Pi}{6}+\cos\left(2x-\dfrac{\Pi}{6}\right)+\cos\left(2x-\dfrac{\Pi}{6}\right)-\cos\dfrac{\Pi}{6}-1=0\)
\(\Leftrightarrow\cos\left(2x-\dfrac{\Pi}{6}\right)=\dfrac{1}{2}\)
3. \(2\sin^22x-3\sin2x\cos2x+\cos^22x=2\)
\(\Leftrightarrow2\sin^22x-3\sin2x\cos2x+\cos^22x-2\left(\sin^22x+\cos^22x\right)=0\)
\(\Leftrightarrow3\sin2x\cos2x+\cos^22x=0\)
\(\Leftrightarrow\cos2x\left(3\sin2x+\cos2x\right)=0\)
-TH1: ...
- TH2: \(\cos2x=-3\sin2x\) mà \(\cos^22x+\sin^22x=1\) suy ra ...
4. \(4\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x+3\sin^2\dfrac{x}{2}=3\)
\(\Leftrightarrow4\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x+3\sin^2\dfrac{x}{2}-3\left(\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x=0\)
\(\Leftrightarrow\dfrac{1+\cos x}{2}+\dfrac{1}{2}\sin x=0\)
\(\Leftrightarrow\cos x+\sin x=-1\)
\(\sin^2x+\dfrac{3}{2}\cos2x + 5 = 0\)
\(\Leftrightarrow \sin^2x+\dfrac{3}{2}(1-2\sin^2x) + 5 = 0\)
\(\Leftrightarrow \sin^2x=\dfrac{13}{4}\)
Suy ra PT vô nghiệm.
Cách khác chi tiết hơn
Ta đã biết \(\cos 2x = \cos^2 x -\sin^2 x = (1-\sin^2 x)-\sin^2 x = 1-2\sin^2 x\)
Vì vậy \(y = \sin^2 x +(1.5)(1-2\sin^2 x) + 5\)
\(\Rightarrow y = -2\sin^2 x + 6.5\). Bây giờ, khi \(\sin x\in [-1,1]\),\(\sin^2 x \in [0,1]\),vậy \(y \in[ 6,5;7,5]\)
Ta dễ dàng thấy \(y=0\) ko trong khoảng, vậy \(y=0\) ko phải là nghiệm cho \(x\)
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
TH1: Xét cox = 0 ( có p là nghiệm ko)
TH2: Xét \(\cos x\ne0\). Ta chia cả hai vế \(\cos^2x\)
Pt trở thành \(2\tan^2x-4\tan x+4-1\left(1+\tan^2x\right)=0\)
\(\Leftrightarrow\tan^2x-4\tan x+3=0\)
Đặt \(\tan x=t\). Giải pt nữa là xg ạ
\(2sin^2x-4sinx.cosx+4cos^2x=1\)
\(\Leftrightarrow2\left(sin^2x+cos^2x\right)-4sinx.cosx+2cos^2x-1=0\)
\(\Leftrightarrow2-2sin2x+cos2x=0\)
\(\Leftrightarrow2sin2x-cos2x=2\)
\(\Leftrightarrow\sqrt{5}\left(\dfrac{2}{\sqrt{5}}sin2x-\dfrac{1}{\sqrt{5}}cos2x\right)=2\)
\(\Leftrightarrow sin\left(2x-arccos\dfrac{2}{\sqrt{5}}\right)=\dfrac{2}{\sqrt{5}}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-arccos\dfrac{2}{\sqrt{5}}=arcsin\dfrac{2}{\sqrt{5}}+k2\pi\\2x-arccos\dfrac{2}{\sqrt{5}}=\pi-arcsin\dfrac{2}{\sqrt{5}}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}arccos\dfrac{2}{\sqrt{5}}+\dfrac{1}{2}arcsin\dfrac{2}{\sqrt{5}}+k\pi\\x=\dfrac{\pi}{2}+\dfrac{1}{2}arccos\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}arcsin\dfrac{2}{\sqrt{5}}+k\pi\end{matrix}\right.\)