Toán Lớp 11: ( √3-1)sinx – ( √3+1)cosx + √3 – 1 = 0
giúp e vs ạ gấp
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 11: ( √3-1)sinx – ( √3+1)cosx + √3 – 1 = 0
giúp e vs ạ gấp
Comments ( 2 )
\tan \dfrac{{5\pi }}{{12}} = \tan \left( {\dfrac{{2\pi }}{3} – \dfrac{\pi }{4}} \right)\\
= \dfrac{{\tan \dfrac{{2\pi }}{3} – \tan \dfrac{\pi }{4}}}{{1 + \tan \dfrac{{2\pi }}{3}\tan \dfrac{\pi }{4}}} = \dfrac{{ – \sqrt 3 – 1}}{{1 – \sqrt 3 }} = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 – 1}}\\
\left( {\sqrt 3 – 1} \right)\sin x – \left( {\sqrt 3 + 1} \right)\cos x + \sqrt 3 – 1 = 0\\
\Leftrightarrow \sin x – \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 – 1}}\cos x + 1 = 0\\
\Leftrightarrow \sin x – \tan \dfrac{{5\pi }}{{12}}\cos x + 1 = 0\\
\Leftrightarrow \cos \dfrac{{5\pi }}{{12}}\sin x – \sin \dfrac{{5\pi }}{{12}}\cos x + \cos \dfrac{{5\pi }}{{12}} = 0\\
\Leftrightarrow \sin \left( {x – \dfrac{{5\pi }}{{12}}} \right) = – \cos \dfrac{{5\pi }}{{12}}\\
\Leftrightarrow \sin \left( {x – \dfrac{{5\pi }}{{12}}} \right) = – \sin \left( {\dfrac{\pi }{{12}}} \right)\\
\Leftrightarrow \sin \left( {x – \dfrac{{5\pi }}{{12}}} \right) = \sin \left( { – \dfrac{\pi }{{12}}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x – \dfrac{{5\pi }}{{12}} = – \dfrac{\pi }{{12}} + k2\pi \\
x – \dfrac{{5\pi }}{{12}} = \dfrac{{13\pi }}{{12}} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x = \dfrac{{3\pi }}{2} + k2\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$