Bài 1.19 trang 39 SGK Toán 11 tập 1 - Kết nối tri thức

Đề bài

Giải các phương trình sau:

a) \(\sin x = \frac{{\sqrt 3 }}{2}\);               

b) \(2\cos x =  - \sqrt 2 \);                  

c) \(\sqrt 3 \tan \left( {\frac{x}{2} + {{15}^o}} \right) = 1\);          

d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\).

Lời giải chi tiết

a) \(\sin x = \frac{{\sqrt 3 }}{2} \)

\(\Leftrightarrow \sin x = \sin \frac{\pi }{3} \)

\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \pi  - \frac{\pi }{3} + k2\pi }\end{array}} \right.\)

\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \frac{{2\pi }}{3} + k2\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right)\).

b) \(2\cos x =  - \sqrt 2  \)

\(\Leftrightarrow \cos x =  - \frac{{\sqrt 2 }}{2}\)

\(\Leftrightarrow \cos x = \cos \frac{{3\pi }}{4} \)

\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{3\pi }}{4} + k2\pi }\\{x =  - \frac{{3\pi }}{4} + k2\pi }\end{array}\left( {k \in \mathbb{Z}} \right)} \right.\).

c) \(\sqrt 3 \left( {\tan \frac{x}{2} + {{15}^0}} \right) = 1 \)

\(\Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \frac{1}{{\sqrt 3 }} \)

\(\Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{6}\)

\( \Leftrightarrow \frac{x}{2} + \frac{\pi }{{12}} = \frac{\pi }{6} + k\pi \)

\(\Leftrightarrow \frac{x}{2} = \frac{\pi }{{12}} + k\pi \)

\(\Leftrightarrow x = \frac{\pi }{6} + 2k\pi \) \(\left( {k \in \mathbb{Z}} \right)\).

d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5} \)

\(\Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \)

\(\Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \)

\(\Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\) \(\left( {k \in \mathbb{Z}} \right)\).