Giải pt a. Sin2x-2cosx=0 b. tan3x+tanx=0 c. 8cos2x.sin2x.cos4x= căn 2

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\sin 2x – 2\cos x = 0\\
 \Leftrightarrow 2\sin x.\cos x – 2\cos x = 0\\
 \Leftrightarrow 2\cos x.\left( {\sin x – 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin x = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{2} + k2\pi 
\end{array} \right. \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \\
b,\\
\tan 3x + \tan x = 0\\
 \Leftrightarrow \tan 3x =  – \tan x\\
 \Leftrightarrow \tan 3x = \tan \left( { – x} \right)\\
 \Leftrightarrow 3x =  – x + k\pi \\
 \Leftrightarrow 4x = k\pi \\
 \Leftrightarrow x = \dfrac{{k\pi }}{4}\\
c,\\
8\cos 2x.\sin 2x.\cos 4x = \sqrt 2 \\
 \Leftrightarrow 4.\left( {2\cos 2x.\sin 2x} \right).\cos 4x = \sqrt 2 \\
 \Leftrightarrow 4.\sin 4x.\cos 4x = \sqrt 2 \\
 \Leftrightarrow 2.\left( {2\sin 4x.\cos 4x} \right) = \sqrt 2 \\
 \Leftrightarrow \sin 8x = \dfrac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \left[ \begin{array}{l}
8x = \dfrac{\pi }{4} + k2\pi \\
8x = \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{32}} + \dfrac{{k\pi }}{4}\\
x = \dfrac{{3\pi }}{{32}} + \dfrac{{k\pi }}{4}
\end{array} \right.
\end{array}\)