Đặt tanx = t

sin2x = 2sinx.cosx

= 2. tanx . cos²x

= \(2tanx.\dfrac{1}{tan^2x+1}\)

= \(\dfrac{2t}{t^2+1}\)

Vậy ta có phương trình

\(\dfrac{2t}{t^2+1}+4t=\dfrac{9\sqrt{3}}{2}\)

giúp mk với ạ

ĐKXĐ; ...

\(\Leftrightarrow\frac{3}{5}sinx+\frac{4}{5}cosx-1=\frac{1}{5}\left(4tanx-3\right)^2\)

\(\Leftrightarrow sin\left(x+a\right)-1=\frac{1}{5}\left(4tanx-3\right)^2\)

(Trong đó \(a\in\left(0;\pi\right)\) sao cho \(cosa=\frac{3}{5}\))

Do \(\left\{{}\begin{matrix}sin\left(x+a\right)-1\le0\\\left(4tanx-3\right)^2\ge0\end{matrix}\right.\) \(\forall a;x\) nên đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}sin\left(x+a\right)=1\\4tanx-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}3sinx+4cosx=5\\4sinx-3cosx=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}sinx=\frac{3}{5}\\cosx=\frac{4}{5}\end{matrix}\right.\) \(\Rightarrow x=arcsin\left(\frac{3}{5}\right)+k2\pi\)

Ta có: y =  tan²x  - 4tan x+ 1 =( tan x – 2)² – 3 

Do đó,  miny = -3 đạt được khi tan x = 2

Không tông tại max y 

Đáp án B

ĐKXĐ: \(cosx\ne0\Rightarrow x\ne\dfrac{\pi}{2}+k\pi\)

\(\dfrac{1}{2}cos4x+\dfrac{4sinx}{cosx}.cos^2x=m\)

\(\Rightarrow\dfrac{1}{2}cos4x+2sin2x=m\)

\(\Rightarrow\dfrac{1}{2}\left(1-2sin^22x\right)+2sin2x=m\)

\(\Rightarrow-sin^22x+2sin2x+\dfrac{1}{2}=m\) 

Đặt \(sin2x=t\in\left[-1;1\right]\Rightarrow-t^2+2t+\dfrac{1}{2}=m\)

Xét hàm \(f\left(t\right)=-t^2+2t+\dfrac{1}{2}\) trên \(\left[-1;1\right]\)

\(-\dfrac{b}{2a}=1\) ; \(f\left(-1\right)=-\dfrac{5}{2}\) ; \(f\left(1\right)=\dfrac{3}{2}\) \(\Rightarrow-\dfrac{5}{2}\le f\left(t\right)\le\dfrac{3}{2}\)

\(\Rightarrow\) Phương trình đã cho vô nghiệm khi \(\left[{}\begin{matrix}m< -\dfrac{5}{2}\\m>\dfrac{3}{2}\end{matrix}\right.\)

\(\frac{1-sin2x}{1+sin2x}=\frac{sin^2x+cos^2x-2sinx.cosx}{sin^2x+cos^2x+2sinx.cosx}=\frac{\left(sinx-cosx\right)^2}{\left(sinx+cosx\right)^2}\)

\(=\frac{\left[\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\right]^2}{\left[\sqrt{2}.sin\left(x+\frac{\pi}{4}\right)\right]^2}=tan^2\left(\frac{\pi}{4}-x\right)\)

Bạn coi lại đề, vế phải là tan chứ ko phải cot

\(\frac{sin2x-2sinx}{sin2x+2sinx}=\frac{2sinx.cosx-2sinx}{2sinx.cosx+2sinx}=\frac{2sinx\left(cosx-1\right)}{2sinx\left(cosx+1\right)}\)

\(=\frac{cosx-1}{cos+1}=\frac{1-2sin^2\frac{x}{2}-1}{2cos^2\frac{x}{2}-1+2}=\frac{-2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}}=-tan^2\frac{x}{2}\)

b, \(VT=\dfrac{1-sin2x}{1+sin2x}\)

\(=\dfrac{sin^2x+cos^2x-2sinx.cosx}{sin^2x+cos^2x+2sinx.cosx}\)

\(=\dfrac{\left(sinx-cosx\right)^2}{\left(sinx+cosx\right)^2}\)

\(=\dfrac{\left(\dfrac{sinx-cosx}{cosx}\right)^2}{\left(\dfrac{sinx+cosx}{cosx}\right)^2}\)

\(=\dfrac{\left(\dfrac{sinx}{cosx}-1\right)^2}{\left(\dfrac{sinx}{cosx}+1\right)^2}\)

\(=\dfrac{\left(tanx-tan\dfrac{\pi}{4}\right)^2}{\left(1+tanx.tan\dfrac{\pi}{4}\right)^2}\)

\(=tan^2\left(x-\dfrac{\pi}{4}\right)=tan^2\left(\dfrac{\pi}{4}-x\right)=VP\)