## Mn giúp mk câu kia với ạk

Question

Mn giúp mk câu kia với ạk

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10 months 2020-11-21T04:25:22+00:00 1 Answers 64 views 0

$\left[ \begin{array}{l} x = \dfrac{\pi }{4} + k\pi \\ x = \dfrac{\pi }{3} + k\pi \end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)$
$$\begin{array}{l} \sin x.\cos y = \dfrac{1}{2}.\left[ {\sin \left( {x + y} \right) + \sin \left( {x – y} \right)} \right]\\ 2\sqrt 2 \sin \left( {x – \dfrac{\pi }{{12}}} \right).\cos x = 1\\ \Leftrightarrow \sqrt 2 .\left[ {2\sin \left( {x – \dfrac{\pi }{{12}}} \right).\cos x} \right] = 1\\ \Leftrightarrow \sqrt 2 .\left[ {\sin \left( {x – \dfrac{\pi }{{12}} + x} \right) + \sin \left( {x – \dfrac{\pi }{{12}} – x} \right)} \right] = 1\\ \Leftrightarrow \sqrt 2 .\left[ {\sin \left( {2x – \dfrac{\pi }{{12}}} \right) + \sin \left( { – \dfrac{\pi }{{12}}} \right)} \right] = 1\\ \Leftrightarrow \sin \left( {2x – \dfrac{\pi }{{12}}} \right) + \dfrac{{ – \sqrt 6 + \sqrt 2 }}{4} = \dfrac{1}{{\sqrt 2 }}\\ \Leftrightarrow \sin \left( {2x – \dfrac{\pi }{{12}}} \right) = \dfrac{{\sqrt 6 + \sqrt 2 }}{4}\\ \Leftrightarrow \left[ \begin{array}{l} 2x – \dfrac{\pi }{{12}} = \dfrac{{5\pi }}{{12}} + k2\pi \\ 2x – \dfrac{\pi }{{12}} = \dfrac{{7\pi }}{{12}} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{4} + k\pi \\ x = \dfrac{\pi }{3} + k\pi \end{array} \right.\,\,\,\,\,\left( {k \in Z} \right) \end{array}$$