Bài tập 28 trang 41 SGK Toán 11 NC

Bài tập 28 trang 41 SGK Toán 11 NC

a) Đặt \(t = \cos x,\left| t \right| \le 1\), ta có:

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

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{\cos ^2}x + \sin x + 1 = 0\\
 \Leftrightarrow 1 - {\sin ^2}x + \sin x + 1 = 0
\end{array}\\
{ \Leftrightarrow {{\sin }^2}x - \sin x - 2 = 0}\\
\begin{array}{l}
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\sin x =  - 1}\\
{\sin x = 2\left( l \right)}
\end{array}} \right.\\
 \Leftrightarrow x =  - \frac{\pi }{2} + k2\pi 
\end{array}
\end{array}\)

c)

\(\begin{array}{l}
\sqrt 3 {\tan ^2}x - \left( {1 + \sqrt 3 } \right)\tan x + 1 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x = \frac{1}{{\sqrt 3 }}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{4} + k\pi \\
x = \frac{\pi }{6} + k\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)

-- Mod Toán 11