Đáp án:
Giải thích các bước giải:
`sin(x) = 1`
Vì `sin(t) = 1` với `t = \pi/2 + 2k\pi` nên `x = \pi/2 + 2k\pi` do đó `x = \pi/2 + 2k\pi` `(k\inZZ)`
`⇔ x = \pi/2 + 2k\pi` `(k\inZZ)`
Vậy `S = {\pi/2 + 2k\pi|k\inZZ}`
`sin(x) = -1`
Vì `sin(t) = -1` với `t = (3\pi)/2 + 2k\pi` nên `x = (3\pi)/2 + 2k\pi` do đó `x = (3\pi)/2 + 2k\pi` `(k\inZZ)`
`⇔ x = (3\pi)/2 + 2k\pi` `(k\inZZ)`
Vậy `S = {(3\pi)/2+2k\pi|k\inZZ}`
`sin(x) = -1/2`
`⇔`\(\left[ \begin{array}{l}\sin(x)=-\dfrac12\\\sin(\pi-x)=-\dfrac12\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-\dfrac\pi6\\\pi-x=\arcsin(-\dfrac12)\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-\dfrac\pi6\\\pi-x=-\dfrac\pi6\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-\dfrac\pi6+2k\pi\\\pi-x=-\dfrac\pi6+2k\pi\end{array} \right.\) `(k\inZZ)`
`⇔`\(\left[ \begin{array}{l}x=\dfrac{11\pi}6+2k\pi\\x=\dfrac{7\pi}6-2k\pi\end{array} \right.\) `(k\inZZ)`
Vậy `S = {(11\pi)/6+2k\pi,(7\pi)/6+2k\pi|k\inZZ}`
