Bài tập 1.25 trang 24 SBT Toán 11 Tập 1 Kết nối tri thức - KNTT

a) Ta có: \(2\sin \left( {\frac{x}{3} + {{15}^0}} \right) + \sqrt 2 = 0 \Leftrightarrow \sin \left( {\frac{x}{3} + {{15}^0}} \right) = \frac{{ - \sqrt 2 }}{2} \Leftrightarrow \sin \left( {\frac{x}{3} + {{15}^0}} \right) = \sin \left( { - {{45}^0}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}\frac{x}{3} + {15^0} = - {45^0} + k{360^0}\\\frac{x}{3} + {15^0} = {180^0} + {45^0} + k{360^0}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - {180^0} + k{1080^0}\\x = {630^0} + k{1080^0}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

b) Ta có: \(\cos \left( {2x + \frac{\pi }{5}} \right) = - 1 \Leftrightarrow \cos \left( {2x + \frac{\pi }{5}} \right) = \cos \pi \\\Leftrightarrow 2x + \frac{\pi }{5} = \pi + k2\pi \\\Leftrightarrow x = \frac{{2\pi }}{5} + k\pi \)

c) Ta có: \(3\tan 2x + \sqrt 3 = 0 \Leftrightarrow \tan 2x = \frac{{ - \sqrt 3 }}{3} \\\Leftrightarrow \tan 2x = \tan \left( {\frac{{ - \pi }}{6}} \right) \Leftrightarrow 2x = \frac{{ - \pi }}{6} + k\pi \left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow x = \frac{{ - \pi }}{{12}} + \frac{{k\pi }}{2}\left( {k \in \mathbb{Z}} \right)\)

d) Ta có: \(\cot \left( {2x - 3} \right) = \cot {15^0} \Leftrightarrow 2x - 3 = {15^0} + k{180^0} \\\Leftrightarrow x = \frac{{{{15}^0} + 3}}{2} + k{90^0} \left( {k \in \mathbb{Z}} \right)\)

-- Mod Toán 11 HỌC247