Toán Lớp 11: 1. Giải Phương trình tan^2x=1-cosx/1-sinx

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Toán Lớp 11:  1. Giải Phương trình tan^2x=1-cosx/1-sinx

trucoanh55 · Answer

Giải đáp:   $ left[ begin{array}{l} x=k 2  pi(k  in  mathbb{Z})    x= dfrac{ pi}{4}+k   pi (k  in  mathbb{Z})  end{array}  right..$   Lời giải và giải thích chi tiết:   $ tan^2x= dfrac{1- cos x}{1- sin x}    text{ĐKXĐ: } left { begin{array}{l} x  ne  dfrac{ pi}{2}+ k  pi(k  in  mathbb{Z} )    sin x  ne 1 end{array}  right.    Leftrightarrow  left { begin{array}{l} x  ne  dfrac{ pi}{2}+ k  pi(k  in  mathbb{Z})    x  ne  dfrac{ pi}{2}+ k2  pi(k  in  mathbb{Z}) end{array}  right.    Leftrightarrow x  ne  dfrac{ pi}{2}+ k  pi(k  in  mathbb{Z})    tan^2x= dfrac{1- cos x}{1- sin x}    Leftrightarrow  tan^2x= dfrac{(1- cos x)(1+ sin x)}{(1- sin x)(1+ sin x)}    Leftrightarrow  tan^2x= dfrac{ sin x- cos x- sin x  cos x+1}{ cos^2x}    Leftrightarrow  sin^2x= sin x- cos x- sin x  cos x+1    Leftrightarrow  sin^2x- sin x+ cos x+ sin x  cos x-1    Leftrightarrow - cos^2x- sin x+ cos x+ sin x  cos x=0    Leftrightarrow - cos^2x+ cos x- sin x+ sin x  cos x=0    Leftrightarrow  cos x(1- cos x+)- sin x(1- cos x)=0    Leftrightarrow (1- cos x)( cos x- sin x)=0    Leftrightarrow  left[ begin{array}{l}  cos x=1     cos x= sin x end{array}  right.    Leftrightarrow  left[ begin{array}{l} x=k 2  pi(k  in  mathbb{Z})     cos x= cos  left( dfrac{ pi}{2}-x right) end{array}  right.     Leftrightarrow  left[ begin{array}{l} x=k 2  pi(k  in  mathbb{Z})    x= dfrac{ pi}{2}-x+k 2  pi (k  in  mathbb{Z})    x=x- dfrac{ pi}{2}+k 2  pi (k  in  mathbb{Z}) end{array}  right.    Leftrightarrow  left[ begin{array}{l} x=k 2  pi(k  in  mathbb{Z})    2x= dfrac{ pi}{2}+k 2  pi (k  in  mathbb{Z})    0=- dfrac{ pi}{2}+k 2  pi (k  in  mathbb{Z})(L) end{array}  right.    Leftrightarrow  left[ begin{array}{l} x=k 2  pi(k  in  mathbb{Z})    x= dfrac{ pi}{4}+k   pi (k  in  mathbb{Z})  end{array}  right..$