phương trình lượng giác

H

hthtb22

$$ \sin \dfrac{x}{2}.\sin x - \cos \dfrac{x}{2}.\sin^2 x+1- 2\cos^2( \dfrac{\pi}{4} - \dfrac{x}{2}) =0$$
phương trình tương đương với:
$$ \sin \dfrac{x}{2}.\sin x - \cos \dfrac{x}{2}.\sin^2 x+1-( \cos \dfrac{x}{2}+ \sin \dfrac{x}{2})^2=0$$$$ \Longleftrightarrow \sin \dfrac{x}{2}.\sin x - \cos \dfrac{x}{2}.\sin^2 x -2\sin \dfrac{x}{2}\cos\dfrac{x}{2}=0$$$$ \Longleftrightarrow \sin \dfrac{x}{2}.\sin x - \cos \dfrac{x}{2}.\sin^2 x -\sin x=0$$$$ \Longleftrightarrow \left[\begin{array}{1} \sin x=0 \\ \sin \dfrac{x}{2} - \cos \dfrac{x}{2}.\sin x -1=0 \end{array}\right.$$$$ \Longleftrightarrow \left[\begin{array}{1} \sin x=0 \\ \sin \dfrac{x}{2} - 2\cos^2 \dfrac{x}{2}.\sin \dfrac{x}{2} -1=0 \end{array}\right.$$$$ \Longleftrightarrow \left[\begin{array}{1} \sin x=0 \\ \sin \dfrac{x}{2} - 2(1-\sin^2 \dfrac{x}{2}).\sin \dfrac{x}{2} -1=0 \end{array}\right.$$$$ \Longleftrightarrow \left[\begin{array}{1} \sin x=0 \\ 2\sin^3 \dfrac{x}{2} -\sin \dfrac{x}{2} -1=0 \end{array}\right.$$$$ \Longleftrightarrow \left[\begin{array}{1} \sin x=0 \\ \sin \dfrac{x}{2} =1 \end{array}\right.$$
 
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