đặt t=sinx+cosx và phải có đk của t

Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=t\) \(\Rightarrow2sinx.cosx=t^2-1\)

Do \(x\in\left[0;\frac{\pi}{2}\right]\Rightarrow x+\frac{\pi}{4}\in\left[\frac{\pi}{4};\frac{3\pi}{4}\right]\) \(\Rightarrow\frac{\sqrt{2}}{2}\le sin\left(x+\frac{\pi}{4}\right)\le1\)

\(\Rightarrow1\le t\le\sqrt{2}\)

Pt trở thành: \(m\left(t+1\right)=t^2\Leftrightarrow m=\frac{t^2}{t+1}\)

Xét \(f\left(t\right)=\frac{t^2}{t+1}\) trên \(\left[1;\sqrt{2}\right]\)

Có \(f\left(t\right)-\frac{1}{2}=\frac{t^2}{t+1}-\frac{1}{2}=\frac{\left(t-1\right)\left(2t+1\right)}{2\left(t+1\right)}\ge0\Rightarrow f\left(t\right)\ge\frac{1}{2}\)

\(f\left(t\right)-2\sqrt{2}+2=\frac{t^2}{t+1}-2\sqrt{2}+2=\frac{\left(t-\sqrt{2}\right)\left(t+2-\sqrt{2}\right)}{t+1}\le0\Rightarrow f\left(t\right)\le2\sqrt{2}-2\)

\(\Rightarrow\frac{1}{2}\le m\le2\sqrt{2}-2\)

d la sai

\(DK:0< x< 10\)

\(\Leftrightarrow\left(2\sin x.\cos x-\cos x\right)+\left(6\sin x-3\right)=0\)

\(\Leftrightarrow\cos x\left(2\sin x-1\right)+3\left(2\sin x-1\right)=0\)

\(\Leftrightarrow\left(2\sin x-1\right)\left(\cos x+3\right)=0\)

\(\Leftrightarrow\sin x=\frac{1}{2}\)

\(\Leftrightarrow x=30\left(l\right)\)

Vay PT voi \(x\in\left(0;10\right)\)vo nghiem

a ) \(2cosx-3sinx+2=0\) 

\(\Leftrightarrow2cosx-3sinx=-2\)  

\(\Leftrightarrow\dfrac{2}{\sqrt{13}}cosx-\dfrac{3}{\sqrt{13}}sinx=-\dfrac{2}{\sqrt{13}}\) 

Thấy : \(\left(\dfrac{2}{\sqrt{13}}\right)^2+\left(\dfrac{-3}{\sqrt{13}}\right)^2=1\) nên tồn tại \(\alpha\) t/m : 

\(sin\alpha=\dfrac{2}{\sqrt{13}};cos\alpha=\dfrac{-3}{\sqrt{13}}\) . . Khi đó : \(sin\alpha.cosx+cos\alpha.sinx=\dfrac{-2}{\sqrt{13}}\)

\(\Leftrightarrow sin\left(\alpha+x\right)=\dfrac{-2}{\sqrt{13}}\) ( p/t cơ bản ) 

b ) \(\dfrac{1+sinx}{1+cosx}=\dfrac{1}{2}\) ( ĐK : \(cosx\ne-1\Leftrightarrow x\ne\left(2k+1\right)\pi\) ; ( k thuộc Z )  ) 

\(\Leftrightarrow2+2sinx=cosx+1\) \(\Leftrightarrow cosx-2sinx=1\) 

Làm giống như a )  

c ) \(cos\left(2x-15^o\right)+sin\left(2x-15^o\right)=-1\)  

Đặt \(t=2x-15^o\) ; ta có : \(cos t + sin t = -1\) 

\(\Leftrightarrow\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)=-1\) \(\Leftrightarrow sin\left(t+\dfrac{\pi}{4}\right)=sin\left(-\dfrac{\pi}{4}\right)\)  

Xong rồi bn làm tiếp ; chú ý đổi ra độ 

\(y=\sqrt{3}cos2x+2sinxcosx-2\)

\(=\sqrt{3}cos2x+sin2x-2\)

Ta có: \(\left|\sqrt{3}cos2x+sin2x\right|\le\sqrt{\left(\sqrt{3}\right)^2+1^2}=2\)

Do đó \(-2\le\sqrt{3}cos2x+sin2x\le2\)

\(\Leftrightarrow-4\le\sqrt{3}cos2x+sin2x-2\le2\).

Ta có: \(\left|\sqrt{3}cosx-sinx\right|\le\sqrt{\left(\sqrt{3}\right)^2+\left(-1\right)^2}=2\)

Do đó \(-2\le\sqrt{3}cosx-sinx\le2\)

1.

\(sin^3x+cos^3x=1-\dfrac{1}{2}sin2x\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)=1-sinx.cosx\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)=1-sinx.cosx\)

\(\Leftrightarrow\left(1-sinx.cosx\right)\left(sinx+cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx.cosx=1\\sinx+cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=2\left(vn\right)\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

2.

\(\left|cosx-sinx\right|+2sin2x=1\)

\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)

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

\(\Leftrightarrow\left|cosx-sinx\right|\left(1-\left|cosx-sinx\right|\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=k\pi\\cos^2x+sin^2x-2sinx.cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)

3.

\(2sin2x-3\sqrt{6}\left|sinx+cosx\right|+8=0\)

\(\Leftrightarrow2\left(sinx+cosx\right)^2-3\sqrt{6}\left|sinx+cosx\right|+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|sinx+cosx\right|=\sqrt{6}\left(vn\right)\\\left|sinx+cosx\right|=\dfrac{\sqrt{6}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left|sin\left(x+\dfrac{\pi}{4}\right)\right|=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\pm\dfrac{\sqrt{3}}{2}\)

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