giải phương trình lượng giác

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xuanquynh97

Bài 1: ĐK: $cosx \not=0$

PT \Leftrightarrow $2(sinx^2-\dfrac{1}{2}(sinx-cosx)^2)=\dfrac{sinx}{cosx}$

\Leftrightarrow $2(sin^2x+2sinxcosx-\dfrac{1}{2})=\dfrac{sinx}{cosx}$

\Leftrightarrow $2sinx(sinx+cosx)-1-\dfrac{sinx}{cosx}$=0

\Leftrightarrow $2sinx(sinx+cosx)-\dfrac{sinx+cosx}{cosx}=0$
 
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xuanquynh97

Bài 2 PT \Leftrightarrow $5cos(3x+\dfrac{\pi}{2})+3cos(5x-\dfrac{\pi}{2})=0$

\Leftrightarrow $3sin5x-5sin3x=0$

\Leftrightarrow $(3sin3x+3sin5x)+2sin3x=0$

\Leftrightarrow $6sin4xcosx+2sin3x=0$

\Leftrightarrow $6sin2xcos2xcosx+sin3x=0$

\Leftrightarrow $12sinxcos^2(x)cos2x+3sinx-4sin^3(x)$

\Leftrightarrow $sinx(12cos^2(x)cos2x+3-4sin^2(x))=0$

PT sau $12((1-sin^2(x))(1-2sin^2(x))+3-4sin^2(x)=0$
 
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