cosx – √3 sinx = √2 3sin ²x + 5cos ²x – 2cos2x – 4sin2x = 0

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1. Đáp án:

    

   Giải thích các bước giải:

   Phương trình tương đương :3sin2x+5cos2x−2(cos2x−sin2x)−8sinx.cosx=03sin2x+5cos2x−2(cos2x−sin2x)−8sinx.cosx=0

   ⇔4sin2x+3cosx2−8sinx.cosx=0⇔4sin2x+3cosx2−8sinx.cosx=0

   ⇔(2sinx−3cosx)(2sinx−cosx)=0

   cosx + √3.sinx = √2
   <=>1/2cosx+√3/2 sinx=√2/2
   <=>cos π/3 cosx+sin π/3 sinx=√2/2
   <=>cos(x- π/3)=cos π/4
   <=>[x=7π/12+k2π
         [x=π/12+k2π

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2. Đáp án:

   $\begin{array}{l}
    + )\\
   \cos x – \sqrt 3 \sin x = \sqrt 2 \\
    \Rightarrow \frac{1}{2}\cos x – \frac{{\sqrt 3 }}{2}\sin x = \frac{{\sqrt 2 }}{2}\\
    \Rightarrow \cos \frac{\pi }{3}.\cos x – \sin \frac{\pi }{3}.\sin x = \frac{{\sqrt 2 }}{2}\\
    \Rightarrow \cos \left( {x + \frac{\pi }{3}} \right) = \cos \frac{\pi }{4}\\
    \Rightarrow \left[ \begin{array}{l}
   x + \frac{\pi }{3} = \frac{\pi }{4} + k2\pi \\
   x + \frac{\pi }{3} =  – \frac{\pi }{4} + k2\pi 
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   x =  – \frac{\pi }{{12}} + k2\pi \\
   x =  – \frac{{7\pi }}{{12}} + k2\pi 
   \end{array} \right.\\
    + )\\
   3{\sin ^2}x + 5{\cos ^2}x – 2\cos 2x – 4\sin 2x = 0\\
    \Rightarrow 3{\sin ^2}x + 5{\cos ^2}x – 2\left( {{{\cos }^2}x – {{\sin }^2}x} \right) – 4\sin 2x = 0\\
    \Rightarrow 5{\sin ^2}x + 3{\cos ^2}x – 4.2\sin x.\cos x = 0\\
    \Rightarrow 5.\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} – 8.\frac{{\sin x}}{{\cos x}} + 3 = 0\left( {do:{{\cos }^2}x \ne 0} \right)\\
    \Rightarrow 5{\tan ^2}x – 8.\tan x + 3 = 0\\
    \Rightarrow \left[ \begin{array}{l}
   \tan x = 1\\
   \tan x = \frac{3}{5}
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   x = \frac{\pi }{4} + k\pi \\
   x = \arctan \frac{3}{5} + k\pi 
   \end{array} \right.
   \end{array}$

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