giải hộ mình vs a)2cos2x-2(√3+1)cosx+2+√3=0 b)√3. tan^2.x – ( 1-√3).tanx-1 = 0 c)tan^2.x+( √3 -1).tanx-√3 = 0 d)cos^2. 2x + sin4x-3^2 .2x = 0 e)3sin^2

giải hộ mình vs
a)2cos2x-2(√3+1)cosx+2+√3=0
b)√3. tan^2.x – ( 1-√3).tanx-1 = 0
c)tan^2.x+( √3 -1).tanx-√3 = 0
d)cos^2. 2x + sin4x-3^2 .2x = 0
e)3sin^2. x.- sinxcosx – 2cos^2.x = 3

0 bình luận về “giải hộ mình vs a)2cos2x-2(√3+1)cosx+2+√3=0 b)√3. tan^2.x – ( 1-√3).tanx-1 = 0 c)tan^2.x+( √3 -1).tanx-√3 = 0 d)cos^2. 2x + sin4x-3^2 .2x = 0 e)3sin^2”

  1. b) \(\sqrt 3 {\tan ^2}x – \left( {1 – \sqrt 3 } \right)\tan x – 1 = 0\)

    \( \Leftrightarrow \left[ \begin{array}{l}\tan x =  – 1\\\tan x = \dfrac{1}{{\sqrt 3 }}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  – \dfrac{\pi }{4} + k\pi \\x = \dfrac{\pi }{6} + k\pi \end{array} \right.\)

    c) \({\tan ^2}x + \left( {\sqrt 3 – 1} \right)\tan x – \sqrt 3 = 0\)

    \( \Leftrightarrow \left[ \begin{array}{l}\tan x = 1\\\tan x =  – \sqrt 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{4} + k\pi \\x =  – \dfrac{\pi }{3} + k\pi \end{array} \right.\)

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