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: Giải phương trình sau : sin^2x + sinxcosx =1/2

Toán Lớp 11: Giải phương trình sau : sin^2x + sinxcosx =1/2

Comments ( 2 )

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

  2. Giải đáp:
    sin^2x + sinxcosx =1/2
    <=> -1/2 + cosx sinx + sin^2x = 0
    <=> 1/2 (-1 + 2 cosx sinx + 2 sin^2x) = 0
    <=> -1 + 2 cosx sinx + 2 sin^2x = 0
    <=> -sqrt(2) sin(-2 x + π/4) = 0
    <=> sin(-2 x + π/4) = 0
    <=> -2 x + π/4 = kπ (k \in \mathbb Z)
    <=> -2 x = – π/4+kπ (k \in \mathbb Z)
    <=> x = π/8-(kπ)/2 (k \in \mathbb Z)
    Vậy phương trình có nghiệm x = π/8-(kπ)/2 (k \in \mathbb Z)

Leave a reply

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

About Thanh Thu