Question

Đã trả lời bởi chuyên gia

Answer 1

`a) cos(x- \pi/3) = sin` {5\pi}/7`

`<=>cos(x- \pi/3)= cos(\pi/2-{5\pi}/7)`

`<=>cos(x- \pi/3)=cos({-3\pi}/{14})`

`<=>`\(\left[ \begin{array}{l}x-\dfrac{\pi}{3}=\dfrac{-3\pi}{14}+k2\pi\\x-\dfrac{\pi}{3}=\dfrac{3\pi}{14}+k2\pi\end{array} \right.\)

`<=>` \(\left[ \begin{array}{l}x=\dfrac{5\pi}{42}+k2\pi\\x=\dfrac{23\pi}{42}+k2\pi\end{array} \right.\)

Vậy, `S= {\frac{5\pi}{42}+k2\pi ; \frac{23\pi}{42}+k2\pi | k ∈ Z }`

`b)tan(x/6+ {2\pi}/5)= 1/3`

`<=>x/6+ {2\pi}/5= arctan``1/3+k\pi,k∈Z`

`<=>x/6={-2\pi}/5+arctan``1/3+k\pi,k∈Z`

`<=>x={-12\pi}/5+6arctan``1/3+6k\pi,k∈Z`

Vậy `S={{-12\pi}/5+6arctan``1/3+6k\pi | k∈Z}`

`c)cot(3x+1)=3`

`<=>3x+1=`arccot`3+k\pi,k∈Z`

`<=>3x=-1+`arccot`3+k\pi,k∈Z`

`<=>x={-1}/3+1/3` arccot`3+{k\pi}/3,k∈Z`

Vậy `S={{-1}/3+1/3` arccot`3+{k\pi}/3,k∈Z}`

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Answer 2