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a) $cotx +1 -cos2x(1+\frac{1}{sinx})=0$$\Leftrightarrow \frac{cosx+sinx}{sinx}-(cosx+sinx)(cosx-sinx).\frac{sinx+1}{sinx}=0$$\Leftrightarrow \frac{cosx+sinx}{sinx}.[1-(cosx-sinx)(sinx+1)]=0$$\Leftrightarrow \frac{sinx+cosx}{sinx}=0 \Leftrightarrow sinx+cosx=0 \Leftrightarrow x= -\frac{\pi}4+k2\pi ; x=\frac{3\pi}4 +k2\pi$Hoặc $1- (cosx-sinx)(sinx+1)=0$$\Leftrightarrow 1+sin^2x+sinx-sinx.conx-cosx=0$$\Leftrightarrow (1+sinx)(1-cosx)+sin^2x$Có $-1 \leq sinx \Rightarrow sinx +1 \geq 0$$cosx \leq 1 \Rightarrow 1- cosx \geq 0$$\Rightarrow (sinx+1)(1-cosx) \geq 0$Mà $sin^2x \geq 0 \Rightarrow (sinx+1)(1-cosx)+sin^2x \geq 0$$\Rightarrow (1+sinx)(1-cosx) +sin^2 =0$$\Leftrightarrow \left\{ \begin{array}{l} 1+sinx=0\\1-cosx=0\\ sin^2x=0 \end{array} \right.$ (hệ PT vô nghiệm)Vậy PT có nghiệm là $x=\frac{\pi}4+k2\pi$ hoặc $x=\frac{3\pi}4 + k2\pi$

a) $cotx +1 -cos2x(1+\frac{1}{sinx})=0$ $\Leftrightarrow \frac{cosx+sinx}{sinx}-(cosx+sinx)(cosx-sinx).\frac{sinx+1}{sinx}=0$$\Leftrightarrow \frac{cosx+sinx}{sinx}.[1-(cosx-sinx)(sinx+1)]=0$$\Leftrightarrow \frac{sinx+cosx}{sinx}=0 \Leftrightarrow sinx+cosx=0 $$\Leftrightarrow x= -\frac{\pi}4+k2\pi ; x=\frac{3\pi}4 +k2\pi$Hoặc$1- (cosx-sinx)(sinx+1)=0 $$\Leftrightarrow 1+sin^2x+sinx-sinx.conx-cosx=0$$ \Leftrightarrow (1+sinx)(1-cosx)+sin^2x$Có$-1 \leq sinx \Rightarrow sinx +1 \geq 0 $$cosx \leq 1 \Rightarrow 1- cosx \geq 0$$ \Rightarrow (sinx+1)(1-cosx) \geq 0$Mà$sin^2x \geq 0 \Rightarrow (sinx+1)(1-cosx)+sin^2x \geq 0 $$\Rightarrow (1+sinx)(1-cosx) +sin^2 =0$$ \Leftrightarrow \left\{ $$\begin{array}{l} 1+sinx=0\\1-cosx=0\\ sin^2x=0 \end{array}$$ \right.$(hệ PT vô nghiệm)Vậy PT có nghiệm là$x=\frac{\pi}4+k2\pi$hoặc$x=\frac{3\pi}4 + k2\pi$