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.

222-9+11+12:2*14+14 = ? ( )

Toán Lớp 11: Sin ( 2x + 9????/2 ) – 3cos ( x-15????/2 ) = 1+2sinx Mong mọi người giải giúp vs ạ

Toán Lớp 11: Sin ( 2x + 9????/2 ) – 3cos ( x-15????/2 ) = 1+2sinx
Mong mọi người giải giúp vs ạ

Comments ( 2 )

  1. Đáp án:
     
    Giải thích các bước giải:
     

    toan-lop-11-sin-2-9-2-3cos-15-2-1-2sin-mong-moi-nguoi-giai-giup-vs-a

  2. $\begin{array}{l} \sin \left( {x + \dfrac{{9\pi }}{2}} \right) – 3\cos \left( {x – \dfrac{{15\pi }}{2}} \right) = 1 + 2\sin x\\  \Leftrightarrow \sin \left( {2x + 4\pi  + \dfrac{\pi }{2}} \right) – 3\cos \left( {x – 8\pi  + \dfrac{\pi }{2}} \right) = 1 + 2\sin x\\  \Leftrightarrow \sin \left( {2x + \dfrac{\pi }{2}} \right) – 3\cos \left( {x + \dfrac{\pi }{2}} \right) = 1 + 2\sin x\\  \Leftrightarrow \cos 2x – 3\sin \left( { – x} \right) = 1 + 2\sin x\\  \Leftrightarrow \cos 2x + 3\sin x = 1 + 2\sin x\\  \Leftrightarrow \cos 2x + \sin x – 1 = 0\\  \Leftrightarrow 1 – 2{\sin ^2}x + \sin x – 1 = 0\\  \Leftrightarrow 2{\sin ^2}x – \sin x = 0\\  \Leftrightarrow \sin x\left( {2\sin x – 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} \sin x = 0\\ \sin x =   \dfrac{1}{2} \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ x = \dfrac{{ \pi }}{6} + k2\pi \\ x = \dfrac{{5\pi }}{6} + k2\pi  \end{array} \right.\left( {k \in \mathbb{Z}} \right) \end{array}$  

Leave a reply

222-9+11+12:2*14+14 = ? ( )