## Mọi người giúp mình mấy bài này với ạ :(

Question

Mọi người giúp mình mấy bài này với ạ 🙁

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4 years 2020-10-19T17:32:11+00:00 1 Answers 62 views 0

1. c)

$$\left\{ \begin{array}{l}sin (\dfrac{π}{6} – \dfrac{x}{3}) \ne 0 \\sin 2x – 1 \ne 0\end{array} \right.$$

<=> $$\left\{ \begin{array}{l}\dfrac{π}{6} – \dfrac{x}{3} \ne kπ\\2x \ne \dfrac{π}{2} + k2π\end{array} \right.$$

<=> $$\left\{ \begin{array}{l}x \ne \dfrac{π}{2} + k3π\\x \ne \dfrac{π}{4} + kπ\end{array} \right.$$ (k ∈ ZZ)

=> D = RR \\ {π/2 + k3π; π/4 + kπ | k ∈ ZZ}

d)

sin 3x – sin(π/3 – x) ne 0

<=> sin 3x ne sin (π/3 – x)

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

<=> $$\left[ \begin{array}{l}x \ne \dfrac{π}{12} + k\dfrac{π}{4}\\x \ne -\dfrac{π}{6} + kπ\end{array} \right.$$ (k ∈ ZZ)

=> D = RR \\ {π/12 + k(π)/4; -π/6 + kπ | k ∈ ZZ}

e)

cos (x/2) – cos 3x ne 0

<=> cos (x/2) ne cos 3x

<=> x/2 ne ±3x + k2π

<=> x ne ±6x + k4π

<=> $$\left[ \begin{array}{l}x \ne k\dfrac{4π}{5}\\x \ne k\dfrac{4π}{7}\end{array} \right.$$

=> D = RR \\ {k(4π)/5; k(4π)/7 | k ∈ ZZ}