\(sin^2x-2m.sinx.cosx-sinx.cosx+2mcos^2x=0\)

\(\Leftrightarrow sinx\left(sinx-cosx\right)-2mcosx\left(sinx-cosx\right)=0\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx-2m.cosx\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=cosx\\sinx=2m.cosx\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=2m\end{matrix}\right.\)

Do \(tanx=1\) ko có nghiệm đã cho nên \(tanx=2m\) phải có nghiệm trên khoảng đã cho

\(\Rightarrow tan\left(\dfrac{\pi}{4}\right)< 2m< tan\left(\dfrac{\pi}{3}\right)\)

\(\Rightarrow1< 2m< \sqrt[]{3}\)

\(\Rightarrow m\in\left(\dfrac{1}{2};\dfrac{\sqrt{3}}{2}\right)\) (hoặc có thể 1 đáp án là tập con của tập này cũng được)

sin x + cos x = 1 + sin x.cos x

⇔ sin x.cos x – sin x – cos x + 1 = 0

⇔ (sinx. cosx –sinx)- (cosx -1 ) =0

⇔ sinx. (cosx – 1) – (cosx -1) = 0

⇔ (sin x – 1)(cos x – 1) = 0

Vậy phương trình có tập nghiệm 

\(M=sinx.cosx+\dfrac{sin^2x}{1+cotx}+\dfrac{cos^2x}{1+tanx}\)

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

\(=sinx.cosx+\dfrac{sin^3x+cos^3x}{cosx+sinx}\)

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

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

\(=sin^2x+cos^2x=1\)

a)

\((\sin x+\cos x)^2=\sin ^2x+2\sin x\cos x+\cos ^2x\)

\(=(\sin ^2x+\cos ^2x)+2\sin x\cos x=1+2\sin x\cos x\)

b)

\(\sin ^4x+\cos ^4x=\sin ^4x+2\sin ^2x\cos ^2x+\cos ^4x-2\sin ^2\cos ^2x\)

\(=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x\)

\(=1-2\sin ^2x\cos ^2x\)

c)

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

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

\(=\sin ^2x\left(\frac{\sin x}{\cos x}\right)^2=\sin ^2x\tan ^2x\)

d)

\(\sin ^6x+\cos ^6x=(\sin ^2x)^3+(\cos ^2x)^3\)

\(=(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)\)

\(=\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x\)

\(=(\sin ^4x+\cos ^4x)-\sin ^2x\cos ^2x=1-2\sin ^2x\cos ^2x-\sin ^2x\cos ^2x\)

\(=1-3\sin ^2x\cos ^2x\) (theo kq phần b)

e)

\(\sin x\cos x(1+\tan x)(1+\cot x)=\sin x\cos x(1+\frac{\sin x}{\cos x})(1+\frac{\cos x}{\sin x})\)

\(=\sin x\cos x.\frac{\cos x+\sin x}{\cos x}.\frac{\sin x+\cos x}{\sin x}\)

\(=(\sin x+\cos x)^2=\sin ^2x+\cos ^2x+2\sin x\cos x\)

\(=1+2\sin x\cos x\)

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P/s: Nói chung cứ bám vào công thức \(\sin ^2x+\cos ^2x=1\)

\(\cos x=0\) ko la nghiem cua pt

\(\cos x\ne0\Rightarrow\tan^2x-\sqrt{3}.\tan x=\dfrac{1}{2\cos^2x}=\dfrac{1}{2}\left(1+\tan^2x\right)\)

\(\Leftrightarrow\dfrac{1}{2}\tan^2x-\sqrt{3}\tan x-\dfrac{1}{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\tan x=2-\sqrt{3}\\\tan x=-2-\sqrt{3}\end{matrix}\right.\Leftrightarrow...\)

y ’   =   x sin 2 α   +   cos 2 α .

cotx=2

=>cosx=2*sin x

\(1+cot^2x=\dfrac{1}{sin^2x}\)

=>\(\dfrac{1}{sin^2x}=1+4=5\)

=>\(sin^2x=\dfrac{1}{5}\)

\(B=\dfrac{sin^2x-2\cdot sinx\cdot2\cdot sinx-1}{5\cdot4sin^2x+sin^2x-3}=\dfrac{-3sin^2x-1}{21sin^2x-3}\)

\(=\dfrac{-\dfrac{3}{5}-1}{\dfrac{21}{5}-3}=-\dfrac{8}{5}:\dfrac{6}{5}=-\dfrac{4}{3}\)

\(cotx=2\Rightarrow tanx=\dfrac{1}{2}\)

\(B=\dfrac{sin^2x-2sinx.cosx-1}{5cos^2x+sin^2x-3}\)

\(\Leftrightarrow B=\dfrac{tan^2x-2tanx-\dfrac{1}{cos^2x}}{5+tan^2x-\dfrac{3}{cos^2x}}\)

\(\Leftrightarrow B=\dfrac{tan^2x-2tanx-1-tan^2x}{5+tan^2x-3-3tan^2x}\)

\(\Leftrightarrow B=\dfrac{-2tanx-1}{2-2tan^2x}\)

\(\Leftrightarrow B=\dfrac{-2.\dfrac{1}{2}-1}{2-2.\dfrac{1}{4}}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)

\(I=\int_0^{\pi/2}\sin^2 x.cos^3 xdx=\int_0^{\pi/2}\sin^2 x.(1-\sin^2 x)d(\sin x)=\dfrac{\sin^3 x}{3}\Big|_0^{\pi/2}-\dfrac{\sin^5 x}{5}\Big|_0^{\pi/2}=\dfrac{2}{15}\)

Do đó diện tích hình phẳng là $S=|I|=\dfrac{2}{15}$

cot x=2>0

=>sin x và cosx cùng dấu

=>sinx*cosx>0

\(1+cot^2x=\dfrac{1}{sin^2x}=1+4=5\)

=>sin^2x=1/5

=>cos^2x=4/5

\(B=\dfrac{1}{5}-2\cdot sinx\cdot cosx-\dfrac{1}{5}\cdot\dfrac{4}{5}+\dfrac{1}{5}-3\)

\(=\dfrac{2}{5}-\dfrac{4}{25}-3-2\cdot\dfrac{1}{\sqrt{5}}\cdot\dfrac{2}{\sqrt{5}}\)

\(=\dfrac{10}{25}-\dfrac{4}{25}-\dfrac{75}{25}-2\cdot\dfrac{2}{5}=\dfrac{-69}{25}-\dfrac{4}{5}=\dfrac{-89}{25}\)