Question

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Answer 1

$\sin (x - 1) = \frac{- \sqrt{3}}{2} \sin (x - 1) = \sin (\frac{- π}{3}) x - 1 = \frac{- π}{3} + k 2 π hoặc x - 1 = π + \frac{π}{3} + k 2 π x = 1 - \frac{π}{3} + k 2 π hoặc x = 1 + \frac{4 π}{3} + k 2 π$

$\sin 5 x = \frac{- 1}{2} \sin 5 x = \sin (\frac{- π}{6}) 5 x = \frac{- π}{6} + k 2 π hoặc 5 x = π + \frac{π}{6} + k 2 π x = \frac{- π}{30} + k \frac{2 π}{5} hoặc x = \frac{7 π}{30} + k \frac{2 π}{5}$ ..

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Answer 2

$\sin (x - 1) = \frac{- \sqrt{3}}{2} \sin (x - 1) = \sin (\frac{- π}{3}) x - 1 = \frac{- π}{3} + k 2 π hoặc x - 1 = π + \frac{π}{3} + k 2 π x = 1 - \frac{π}{3} + k 2 π hoặc x = 1 + \frac{4 π}{3} + k 2 π$

$\sin 5 x = \frac{- 1}{2} \sin 5 x = \sin (\frac{- π}{6}) 5 x = \frac{- π}{6} + k 2 π hoặc 5 x = π + \frac{π}{6} + k 2 π x = \frac{- π}{30} + k \frac{2 π}{5} hoặc x = \frac{7 π}{30} + k \frac{2 π}{5}$ ..

Answer 3

`sin(x-1)=-\sqrt{3}/2`

`⇔sin(x-1) = sin\frac{-\pi}{3}`

`⇔`$\left[\begin{matrix} x-1=\frac{-\pi}{3}+2k\pi\\ x-1=\frac{4\pi}{3}+2k\pi\end{matrix}\right.$

`⇔`$\left[\begin{matrix} x=\frac{2\pi}{3}+2k\pi\\ x=\frac{7\pi}{3}+2k\pi\end{matrix}\right.$

Vậy `\S = {\frac{2\pi}{3}+2k\pi ; \frac{7\pi}{3}+2k\pi, k \in \Z}`

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`sin(x-1)=-1/2`

`⇔sin(x-1) = sin{-\frac{\pi}{6}`

`⇔`$\left[\begin{matrix} x-1=\frac{-\pi}{6}+2k\pi\\ x-1=\frac{7\pi}{6}+2k\pi\end{matrix}\right.$

`⇔`$\left[\begin{matrix} x=\frac{5\pi}{6}+2k\pi\\ x=\frac{13\pi}{6}+2k\pi\end{matrix}\right.$

Vậy `\S = {\frac{5\pi}{6}+2k\pi ; \frac{13\pi}{6}+2k\pi, k \in \Z}`

Answer 4

a,

$\sin (x - 1) = \frac{- \sqrt{3}}{2} \Leftrightarrow \sin (x - 1) = \sin (\frac{- π}{3}) \Rightarrow x - 1 = \frac{- π}{3} + k 2 π hoặc (k \in Z) x - 1 = π - \frac{- π}{3} + k 2 π \Leftrightarrow x = \frac{- π}{3} + 1 + k 2 π hoặc (k \in Z) x = \frac{4 π}{3} + 1 + k 2 π b, \sin 5 x = \frac{- 1}{2} \Leftrightarrow \sin 5 x = \sin (\frac{- π}{6}) \Rightarrow 5 x = \frac{- π}{6} + k 2 π hoặc (k \in Z) 5 x = π - \frac{- π}{6} + k 2 π \Leftrightarrow x = \frac{- π}{30} + k \frac{2 π}{5} hoặc (k \in Z) 5 x = \frac{7 π}{6} + k 2 π \Leftrightarrow x = \frac{- π}{30} + k \frac{2 π}{5} hoặc (k \in Z) x = \frac{7 π}{30} + k \frac{2 π}{5}$

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Answer 5

a,

$\sin (x - 1) = \frac{- \sqrt{3}}{2} \Leftrightarrow \sin (x - 1) = \sin (\frac{- π}{3}) \Rightarrow x - 1 = \frac{- π}{3} + k 2 π hoặc (k \in Z) x - 1 = π - \frac{- π}{3} + k 2 π \Leftrightarrow x = \frac{- π}{3} + 1 + k 2 π hoặc (k \in Z) x = \frac{4 π}{3} + 1 + k 2 π b, \sin 5 x = \frac{- 1}{2} \Leftrightarrow \sin 5 x = \sin (\frac{- π}{6}) \Rightarrow 5 x = \frac{- π}{6} + k 2 π hoặc (k \in Z) 5 x = π - \frac{- π}{6} + k 2 π \Leftrightarrow x = \frac{- π}{30} + k \frac{2 π}{5} hoặc (k \in Z) 5 x = \frac{7 π}{6} + k 2 π \Leftrightarrow x = \frac{- π}{30} + k \frac{2 π}{5} hoặc (k \in Z) x = \frac{7 π}{30} + k \frac{2 π}{5}$

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