Giải phương trình: 1)4cot2x-1/sin²x-2=0 2)2cos²2x-sin²x-2=0 3)5(sinx-cosx)²=cos4x+2 4)2sin²xcos²x-sin2x+1/2=0 5)sin²3x+5cos(17π/2-x)+4=0

Giải phương trình:
1)4cot2x-1/sin²x-2=0
2)2cos²2x-sin²x-2=0
3)5(sinx-cosx)²=cos4x+2
4)2sin²xcos²x-sin2x+1/2=0
5)sin²3x+5cos(17π/2-x)+4=0

0 bình luận về “Giải phương trình: 1)4cot2x-1/sin²x-2=0 2)2cos²2x-sin²x-2=0 3)5(sinx-cosx)²=cos4x+2 4)2sin²xcos²x-sin2x+1/2=0 5)sin²3x+5cos(17π/2-x)+4=0”

  1. Đáp án:

    $\begin{array}{l}
    1)\dfrac{{4\cot 2x – 1}}{{{{\sin }^2}x – 2}} = 0\\
    Dkxd:\left\{ \begin{array}{l}
    {\sin ^2}x \ne 2\left( {luon\,dung} \right)\\
    \sin 2x \ne 0
    \end{array} \right.\\
     \Rightarrow 2x \ne k\pi \\
     \Rightarrow x \ne \dfrac{{k\pi }}{2}\\
    Pt \Rightarrow 4\cot 2x – 1 = 0\\
     \Rightarrow \cot 2x = \dfrac{1}{4}\\
     \Rightarrow 2x = arc\cot \dfrac{1}{4} + k\pi \\
     \Rightarrow x = \dfrac{1}{2}arc\cot \dfrac{1}{4} + \dfrac{{k\pi }}{2}\left( {tmdk} \right)\\
    2)2{\cos ^2}2x – {\sin ^2}x – 2 = 0\\
     \Rightarrow 2.{\left( {1 – 2{{\sin }^2}x} \right)^2} – {\sin ^2}x – 2 = 0\\
     \Rightarrow 2.\left( {4{{\sin }^4}x – 4{{\sin }^2}x + 1} \right) – {\sin ^2}x – 2 = 0\\
     \Rightarrow 8{\sin ^4}x – 9{\sin ^2}x = 0\\
     \Rightarrow {\sin ^2}x\left( {8{{\sin }^2}x – 9} \right) = 0\\
     \Rightarrow \left[ \begin{array}{l}
    {\sin ^2}x = 0\\
    {\sin ^2}x = \dfrac{9}{8}\left( {vo\,nghiem} \right)
    \end{array} \right.\\
     \Rightarrow \sin x = 0\\
     \Rightarrow x = k\pi \\
    3)5{\left( {\sin x – \cos x} \right)^2} = \cos 4x + 2\\
     \Rightarrow 5.\left( {{{\sin }^2}x + {{\cos }^2}x – 2\sin x.\cos x} \right) = \cos 4x + 2\\
     \Rightarrow 5.\left( {1 – \sin 2x} \right) = 1 – 2{\sin ^2}2x + 2\\
     \Rightarrow 2{\sin ^2}2x – 5\sin 2x + 2 = 0\\
     \Rightarrow \left( {2\sin 2x – 1} \right)\left( {\sin 2x – 2} \right) = 0\\
     \Rightarrow \sin 2x = \dfrac{1}{2}\\
     \Rightarrow \left[ \begin{array}{l}
    2x = \dfrac{\pi }{6} + k2\pi \\
    2x = \dfrac{{5\pi }}{6} + k2\pi 
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{{12}} + k\pi \\
    x = \dfrac{{5\pi }}{{12}} + k\pi 
    \end{array} \right.\\
    4)2{\sin ^2}x.{\cos ^2}x – \sin 2x + \dfrac{1}{2} = 0\\
     \Rightarrow \dfrac{1}{2}{\sin ^2}2x – \sin 2x + \dfrac{1}{2} = 0\\
     \Rightarrow {\sin ^2}2x – 2\sin x + 1 = 0\\
     \Rightarrow {\left( {\sin 2x – 1} \right)^2} = 0\\
     \Rightarrow \sin 2x = 1\\
     \Rightarrow 2x = \dfrac{\pi }{2} + k2\pi \\
     \Rightarrow x = \dfrac{\pi }{4} + k\pi \\
    5){\sin ^2}3x + 5\cos \left( {\dfrac{{17\pi }}{2} – x} \right) + 4 = 0\\
     \Rightarrow {\sin ^2}3x + 5\sin x + 4 = 0\\
     \Rightarrow 3{\mathop{\rm sinx}\nolimits}  – 4si{n^3}x + 5\sin x + 4 = 0\\
     \Rightarrow 4{\sin ^3}x – 8\sin x – 4 = 0\\
     \Rightarrow {\sin ^3}x – 2\sin x – 1 = 0\\
     \Rightarrow \left( {\sin x + 1} \right)\left( {{{\sin }^2}x – \sin x – 1} \right) = 0\\
     \Rightarrow \left[ \begin{array}{l}
    \sin x =  – 1\\
    \sin x = \dfrac{{1 – \sqrt 5 }}{2}
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    x = \dfrac{{ – \pi }}{2} + k2\pi \\
    x = \arcsin \left( {\dfrac{{1 – \sqrt 5 }}{2}} \right) + k2\pi \\
    x = \pi  – \arcsin \left( {\dfrac{{1 – \sqrt 5 }}{2}} \right) + k2\pi 
    \end{array} \right.
    \end{array}$

    Bình luận

Viết một bình luận