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$sin^3 x+cos^3 x=1$
$\leftrightarrow (sinx +cosx)(sin^2x + cos^2x -sinxcosx )=1$
$\leftrightarrow (sinx+cosx)(1-sinxcosx)=1$
Đặt $sinx+cosx=t=\sqrt{2}sin(x+\dfrac{\pi }{4}) (-\sqrt{2}\le t \le \sqrt{2}) $
$\rightarrow sinxcosx=\dfrac{t^2-1}{2}$
$\rightarrow$ PT $\leftrightarrow t(1-\dfrac{t^2-1}{2})=1$
$\leftrightarrow t^3-3t+2=0$
$\leftrightarrow \begin{bmatrix}& t=-2 & \\ & t=1 & \end{bmatrix}$
$\rightarrow t=1$
$\leftrightarrow \sqrt{2}sin(x+\dfrac{\pi }{4})=1$
$\leftrightarrow sin(x+\dfrac{\pi }{4})=\dfrac{1}{\sqrt{2}}=sin\dfrac{\pi }{4}$
$\rightarrow \begin{bmatrix}& x+\dfrac{\pi }{4}= \dfrac{\pi }{4}+k2\pi & \\ & x+\dfrac{\pi }{4}=\pi -\dfrac{\pi }{4} +k2\pi & \end{bmatrix}$
$\leftrightarrow \begin{bmatrix} & x=k2\pi & \\ & x= \dfrac{\pi }{2} +k2\pi&\end{bmatrix}$
$sinx-cosx+4sinxcosx+1=0$
$sinx-cosx+4sinxcosx+1=0$
Đặt $sinx-cosx=t=\sqrt{2}sin(x-\dfrac{\pi }{4}) (-\sqrt{2}\le t \le \sqrt{2})$
$\rightarrow 4sinxcosx=2-2t^2$
$\rightarrow PT \leftrightarrow t+2-2t^2+1=0$
$\leftrightarrow \begin{bmatrix}& t= \dfrac{3 }{2}& \\ & t=1 & \end{bmatrix}$
$\rightarrow t=1$
$\leftrightarrow \sqrt{2}sin(x-\dfrac{\pi }{4})=1$
$\leftrightarrow sin(x-\dfrac{\pi }{4})=\dfrac{1}{\sqrt{2}}=sin\dfrac{\pi }{4}$
$\rightarrow \begin{bmatrix}& x-\dfrac{\pi }{4}= \dfrac{\pi }{4}+k2\pi & \\ & x-\dfrac{\pi }{4}=\pi -\dfrac{\pi }{4} +k2\pi & \end{bmatrix}$
$\leftrightarrow \begin{bmatrix} & x=\dfrac{\pi }{2}+k2\pi & \\ & x= \pi +k2\pi&\end{bmatrix}$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn