Giải phương trình: 3cotx-2=0 3tanx+4=0 6sinx-3=0 4cotx+2=0 cotx-1=0 sin(x + pi/2) sin 3x=0

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1. Đáp án:

   $\begin{array}{l}
   a)3\cot x – 2 = 0\\
    \Rightarrow \cot x = \dfrac{2}{3}\\
    \Rightarrow x = arc\cot \dfrac{2}{3} + k\pi \\
   b)3\tan x + 4 = 0\\
    \Rightarrow \tan x =  – \dfrac{4}{3}\\
    \Rightarrow x = \arctan \left( { – \dfrac{4}{3}} \right) + k\pi \\
   c)6\sin x – 3 = 0\\
    \Rightarrow \sin x = \dfrac{1}{2}\\
    \Rightarrow \left[ \begin{array}{l}
   x = \dfrac{\pi }{6} + k2\pi \\
   x = \dfrac{{5\pi }}{6} + k2\pi 
   \end{array} \right.\\
   d)4\cot x + 2 = 0\\
    \Rightarrow \cot x =  – \dfrac{1}{2}\\
    \Rightarrow x = ar\cot \left( { – \dfrac{1}{2}} \right) + k\pi \\
   e)\cot x – 1 = 0\\
    \Rightarrow \cot x = 1\\
    \Rightarrow x = \dfrac{\pi }{4} + k\pi \\
   f)sin3x = 0\\
    \Rightarrow 3x = k\pi \\
    \Rightarrow x = \dfrac{{k\pi }}{3}
   \end{array}$

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2. `a) 3cot x – 2 = 0`

   `<=> cot x = 2/3`

   `<=> x = ar“cc“ot (2/3) + kπ, k ∈ ZZ`

   `b) 3tan x + 4 = 0`

   `<=> tan x = -4/3`

   `<=> x = arctan (-4/3) + kπ, k ∈ ZZ`

   `c) 6sin x – 3 = 0`

   `<=> sin x = 1/2`

   `<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{6} + k2π\\x = \dfrac{5π}{6} + k2π\end{array} \right.\) `(k ∈ ZZ)`

   `d) 4cot x + 2 = 0`

   `<=> cot x = -1/2`

   `<=> x = ar“cc“ot (-1/2) + kπ, k ∈ ZZ`

   `e) cot x – 1 = 0`

   `<=> cot x = 1`

   `<=> x = π/4 + kπ, k ∈ ZZ`

   `f) sin (x + π/2) = 0`

   `<=> x + π/2 = kπ`

   `<=> x = -π/2 + kπ, k ∈ ZZ`

   `g) sin 3x = 0`

   `<=> 3x = kπ`

   `<=> x = k(π)/3, k ∈ ZZ`

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