3 sin^2 x – 5 Sin x – 2 = 0 3cos^2 x- sin^2 x – sin2x = 0

3 sin^2 x – 5 Sin x – 2 = 0
3cos^2 x- sin^2 x – sin2x = 0

0 bình luận về “3 sin^2 x – 5 Sin x – 2 = 0 3cos^2 x- sin^2 x – sin2x = 0”

  1. Đáp án:

    $\begin{array}{l}
     + )3{\sin ^2}x – 5\sin x – 2 = 0\\
     \Rightarrow 3{\sin ^2}x – 6\sin x + \sin x – 2 = 0\\
     \Rightarrow \left( {\sin \,x – 2} \right)\left( {3\sin \,x + 1} \right) = 0\\
     \Rightarrow \sin \,x =  – \frac{1}{3}\left( {do:\sin \,x \le 1} \right)\\
     \Rightarrow \left[ \begin{array}{l}
    x = \arcsin \left( { – \frac{1}{3}} \right) + k2\pi \\
    x = \pi  – \arcsin \left( { – \frac{1}{3}} \right) + k2\pi 
    \end{array} \right.\\
     + )3{\cos ^2}x – {\sin ^2}x – \sin 2x = 0\\
     \Rightarrow 3{\cos ^2}x – 2\sin x.\cos x – {\sin ^2}x = 0\\
     \Rightarrow 3{\cos ^2}x – 3\sin x.\cos x + \sin x.\cos x – {\sin ^2}x = 0\\
     \Rightarrow \left( {\cos x – \sin x} \right)\left( {3\cos x + \sin x} \right) = 0\\
     \Rightarrow \left[ \begin{array}{l}
    \cos x = \sin x\\
    3\cos x =  – \sin x
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    \tan x = 1\\
    \tan x =  – 3
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    x = \frac{\pi }{4} + k\pi \\
    x = \arctan \left( { – 3} \right) + k\pi 
    \end{array} \right.
    \end{array}$

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