Bài tập 1.31 trang 38 SBT Toán 11

ĐKXĐ: sinx ≠ 0 và cosx ≠ 0 ⇔ sin2x ≠ 0

⇔ cos2x ≠ ±1

Ta có: \(\cot x - \tan x + 4\sin 2x = \frac{2}{{\sin 2x}}\)

\(\begin{array}{l}
 \Leftrightarrow \frac{{\cos x}}{{\sin x}} - \frac{{\sin x}}{{\cos x}} + 4\sin 2x = \frac{2}{{\sin 2x}}\\
 \Leftrightarrow \frac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sin x\cos x}} + 4\sin 2x = \frac{2}{{\sin 2x}}\\
 \Leftrightarrow \frac{{\cos 2x}}{{\frac{{\sin 2x}}{2}}} + 4\sin 2x = \frac{2}{{\sin 2x}}\\
 \Leftrightarrow \frac{{2\cos 2x}}{{\sin 2x}} + 4\sin 2x = \frac{2}{{\sin 2x}}\\
 \Leftrightarrow 2\cos 2x + 4{\sin ^2}2x = 2\\
 \Leftrightarrow 2\cos 2x + 4(1 - {\cos ^2}2x) = 2\\
 \Leftrightarrow 4{\cos ^2}2x - 2\cos 2x + 2 = 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\cos 2x = 1\,{\rm{(l)}}}\\
{\cos 2x =  - \frac{1}{2}}
\end{array}} \right.\\
 \Leftrightarrow 2x =  \pm \frac{{2\pi }}{3} + k2\pi ,k \in Z\\
 \Leftrightarrow x =  \pm \frac{\pi }{3} + k\pi ,k \in Z
\end{array}\)

-- Mod Toán 11 HỌC247