Bài tập 1.15 trang 23 SBT Toán 11

a) \(\cos (x + 3) = \frac{1}{3}\)

\( \Leftrightarrow x + 3 =  \pm \arccos 13 + k2\pi ,k \in Z\)

\( \Leftrightarrow x =  - 3 \pm \arccos \frac{1}{3} + k2\pi ,k \in Z\)

b) \(\cos (3x - {45^o}) = \frac{{\sqrt 3 }}{2}\)

⇔ cos(3x−45^o) = cos30^o

\( \Leftrightarrow 3x - {45^o} =  \pm {30^o} + k{360^o},k \in Z\)

\( \Leftrightarrow \left[ \begin{array}{l}
x = {25^o} + k{120^o},k \in Z\\
x = {5^o} + k{120^o},k \in Z
\end{array} \right.\)

c) \(\cos (2x + \frac{\pi }{3}) =  - \frac{1}{2}\)

\( \Leftrightarrow \cos (2x + \frac{\pi }{3}) = \cos \left( {\frac{{2\pi }}{3}} \right)\)

\( \Leftrightarrow 2x + \frac{\pi }{3} =  \pm \frac{{2\pi }}{3} + k2\pi ,k \in Z\)

\( \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + k\pi ,k \in Z\\
x =  - \frac{\pi }{2} + k\pi ,k \in Z
\end{array} \right.\)

d) (2+cosx)(3cos2x−1) = 0

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

Nếu cosx = −2 (vô nghiệm)

Nếu \(\cos 2x = \frac{1}{3}\)

\( \Leftrightarrow 2x =  \pm \arccos \frac{1}{3} + k2\pi ,k \in Z\)

\( \Leftrightarrow x =  \pm 12\arccos \frac{1}{3} + k\pi ,k \in Z\)

-- Mod Toán 11 HỌC247