## Giải các phương trình bài 2

Question

Giải các phương trình bài 2

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1 year 2020-11-27T07:29:32+00:00 2 Answers 50 views 0

## Answers ( )

1. a) sin x = 1

<=> x = π/2 + k2π

b) cos 2x = -1

<=> 2x = π + k2π

<=> x = π/2 + kπ

c) tan 3x = 0

<=> 3x = kπ

<=> x = k(π)/2

d) cot 5x = -1

<=> 5x = -π/4 + kπ

<=> x = -π/20 + k(π)/5

e) sin (3x – π/3) = (\sqrt{3})/2

<=> $$\left[ \begin{array}{l}3x – \dfrac{π}{3} = \dfrac{π}{3} + k2π\\3x – \dfrac{π}{3} = \dfrac{2π}{3} + k2π\end{array} \right.$$

<=> $$\left[ \begin{array}{l}x = \dfrac{2π}{9} + k\dfrac{2π}{3}\\x = \dfrac{π}{3} + k\dfrac{2π}{3}\end{array} \right.$$ (k ∈ ZZ)

f) cos (2x – π/4) = (\sqrt{2})/2

<=> $$\left[ \begin{array}{l}2x – \dfrac{π}{4} = \dfrac{π}{4} + k2π\\2x – \dfrac{π}{4} = -\dfrac{π}{4} + k2π\end{array} \right.$$

<=> $$\left[ \begin{array}{l}x = \dfrac{π}{4} + kπ\\x = kπ\end{array} \right.$$ (k ∈ ZZ)

g) tan ((3x)/2 + π/6) = sqrt{3}

<=> (3x)/2 + π/6 = π/3 + kπ

<=> (3x)/2 = π/6 + kπ

<=> x = π/9 + k(2π)/3 (k ∈ ZZ)

2. Đáp án:

$\begin{array}{l} 1)\sin x = 1\\ \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \\ 2)\\ \cos 2x = – 1\\ \Leftrightarrow 2x = \pi + k2\pi \\ \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \\ 3)\tan 3x = 0\\ \Leftrightarrow 3x = k\pi \\ \Leftrightarrow x = \dfrac{{k\pi }}{3}\\ 4)cot5x = – 1\\ \Leftrightarrow 5x = – \dfrac{\pi }{4} + k\pi \\ \Leftrightarrow x = – \dfrac{\pi }{{20}} + \dfrac{{k\pi }}{5}\\ 5)\sin \left( {3x – \dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}\\ \Leftrightarrow \left[ \begin{array}{l} 3x – \dfrac{\pi }{3} = \dfrac{\pi }{3} + k2\pi \\ 3x – \dfrac{\pi }{3} = \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{2\pi }}{9} + \dfrac{{k2\pi }}{3}\\ x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3} \end{array} \right.\\ 6)\cos \left( {2x – \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\\ \Leftrightarrow \left[ \begin{array}{l} 2x – \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\ 2x – \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{4} + k\pi \\ x = \dfrac{\pi }{2} + k\pi \end{array} \right.\\ 7)\tan \left( {\dfrac{{3x}}{2} + \dfrac{\pi }{6}} \right) = \sqrt 3 \\ \Leftrightarrow \dfrac{{3x}}{2} + \dfrac{\pi }{6} = \dfrac{\pi }{3} + k\pi \\ \Leftrightarrow x = \dfrac{\pi }{9} + \dfrac{{k2\pi }}{3} \end{array}$