Toán Lớp 11: a)sin (2x+π) = √2 2 2 b)cos(x-π) = -√3 2 c)tn(2x+π)=-1 3 d) cot( x-π)= -√

Question

Toán Lớp 11:  a)sin (2x+π) = √2                  2       2 b)cos(x-π) = -√3                         2 c)tn(2x+π)=-1                3 d) cot( x-π)= -√3                 3      3

ngoclanquynh · Answer

CH&Uacute;C BẠN HỌC TỐT!!!       Lời giải và giải thích chi tiết:

haiyemy · Answer

~rai~        (a) sin left(2x+ dfrac{ pi}{2} right)= dfrac{ sqrt{2}}{2}   Leftrightarrow  left[ begin{array}{I}2x+ dfrac{ pi}{2}= dfrac{ pi}{4}+k2 pi  2x+ dfrac{ pi}{2}= pi- dfrac{ pi}{4}+k2 pi end{array} right.   Leftrightarrow  left[ begin{array}{I}2x=- dfrac{ pi}{4}+k2 pi  2x= dfrac{ pi}{4}+k2 pi end{array} right.   Leftrightarrow  left[ begin{array}{I}x- dfrac{ pi}{8}+k pi  x= dfrac{ pi}{8}+k pi. end{array} right. quad(k in mathbb{Z})  b) cos(x- pi)=- dfrac{ sqrt{3}}{2}   Leftrightarrow  cos( pi-x)=- dfrac{ sqrt{3}}{2}   Leftrightarrow - cos x=- dfrac{ sqrt{3}}{2}   Leftrightarrow  cos x= dfrac{ sqrt{3}}{2}   Leftrightarrow  left[ begin{array}{I}x= dfrac{ pi}{6}+k2 pi  x=- dfrac{ pi}{6}+k2 pi. end{array} right. quad(k in mathbb{Z})  c) tan left(2x+ dfrac{ pi}{3} right)=-1 quad(1)  ĐKXĐ: cos left(2x+ dfrac{ pi}{3} right) ne 0   Leftrightarrow 2x+ dfrac{ pi}{3} ne dfrac{ pi}{2}+k pi   Leftrightarrow 2x ne dfrac{ pi}{6}+k pi   Leftrightarrow x ne dfrac{ pi}{12}+k dfrac{ pi}{2}.(k in mathbb{Z})  (1) Leftrightarrow 2x+ dfrac{ pi}{3}=- dfrac{ pi}{4}+k pi   Leftrightarrow 2x=- dfrac{7 pi}{12}+k pi   Leftrightarrow x=- dfrac{7 pi}{24}+k dfrac{ pi}{2}.(k in mathbb{Z})  d) cot left(x- dfrac{ pi}{3} right)=- dfrac{ sqrt{3}}{3} quad(1)  ĐKXĐ: sin left(x- dfrac{ pi}{3} right) ne 0   Leftrightarrow x- dfrac{ pi}{3} ne k pi   Leftrightarrow x ne dfrac{ pi}{3}+k pi.(k in mathbb{Z})  (1) Leftrightarrow x- dfrac{ pi}{3}=- dfrac{ pi}{3}+k pi   Leftrightarrow x=k pi.(k in mathbb{Z}) )