Chương 1: HÀM SỐ LƯỢNG GIÁC. PHƯƠNG TRÌNH LƯỢNG GIÁC

TN

Làm giúp mình với ạ, mìn đang cần gấp. Làm từ câu 4 ạ

NL
13 tháng 6 2021 lúc 0:07

Tất cả k dưới đây đều là \(k\in Z\)

6.

\(\Leftrightarrow\sqrt{3}cot\left(3x-\dfrac{\pi}{3}\right)=1\)

\(\Leftrightarrow cot\left(3x-\dfrac{\pi}{3}\right)=\dfrac{1}{\sqrt{3}}\)

\(\Leftrightarrow cot\left(3x-\dfrac{\pi}{3}\right)=cot\left(\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow3x-\dfrac{\pi}{3}=\dfrac{\pi}{3}+k\pi\)

\(\Leftrightarrow3x=\dfrac{2\pi}{3}+k\pi\)

\(\Leftrightarrow x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\) 

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NL
13 tháng 6 2021 lúc 0:09

7.

\(\Leftrightarrow\sqrt{3}tan\left(3x-15^0\right)=-1\)

\(\Leftrightarrow tan\left(3x-15^0\right)=-\dfrac{1}{\sqrt{3}}\)

\(\Leftrightarrow tan\left(3x-15^0\right)=tan\left(-30^0\right)\)

\(\Leftrightarrow3x-15^0=-30^0+k180^0\)

\(\Leftrightarrow3x=-15^0+k180^0\)

\(\Leftrightarrow x=-3^0+k60^0\)

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NL
13 tháng 6 2021 lúc 0:10

8.

\(\Leftrightarrow\sqrt{3}cot\left(3x-30^0\right)=1\)

\(\Leftrightarrow cot\left(3x-30^0\right)=\dfrac{1}{\sqrt{3}}\)

\(\Leftrightarrow cot\left(3x-30^0\right)=cot\left(60^0\right)\)

\(\Leftrightarrow3x-30^0=60^0+k180^0\)

\(\Leftrightarrow3x=90^0+k180^0\)

\(\Leftrightarrow x=30^0+k60^0\)

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NL
13 tháng 6 2021 lúc 0:11

9.

\(\Leftrightarrow sin\left(3x-\dfrac{\pi}{4}\right)=sin\left(\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow3x-\dfrac{\pi}{4}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow3x=\dfrac{3\pi}{4}+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k2\pi}{3}\)

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NL
13 tháng 6 2021 lúc 0:12

10.

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=sin\left(-\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow2x-\dfrac{\pi}{3}=-\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow2x=-\dfrac{\pi}{6}+k2\pi\)

\(\Leftrightarrow x=-\dfrac{\pi}{12}+k\pi\)

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NL
13 tháng 6 2021 lúc 0:13

11.

\(\Leftrightarrow sin\left(3x-\dfrac{\pi}{4}\right)=sin\left(0\right)\)

\(\Leftrightarrow3x-\dfrac{\pi}{4}=k\pi\)

\(\Leftrightarrow3x=\dfrac{\pi}{4}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\)

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NL
13 tháng 6 2021 lúc 0:14

12.

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\left(0\right)\)

\(\Leftrightarrow2x-\dfrac{\pi}{3}=k2\pi\)

\(\Leftrightarrow2x=\dfrac{\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)

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NL
13 tháng 6 2021 lúc 0:15

13.

\(\Leftrightarrow cos\left(3x-\dfrac{\pi}{4}\right)=cos\left(\pi\right)\)

\(\Leftrightarrow3x-\dfrac{\pi}{4}=\pi+k2\pi\)

\(\Leftrightarrow3x=\dfrac{5\pi}{4}+k2\pi\)

\(\Leftrightarrow x=\dfrac{5\pi}{12}+\dfrac{k2\pi}{3}\)

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NL
13 tháng 6 2021 lúc 0:16

14.

\(\Leftrightarrow3x-\dfrac{\pi}{4}=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow3x=\dfrac{3\pi}{4}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{3}\)

15.

\(\Leftrightarrow3x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k\pi\)

\(\Leftrightarrow3x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\)

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NL
13 tháng 6 2021 lúc 0:17

16.

\(\Leftrightarrow2x-\dfrac{\pi}{3}=-\dfrac{\pi}{4}+k\pi\)

\(\Leftrightarrow2x=\dfrac{\pi}{12}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{24}+\dfrac{k\pi}{2}\)

17.

\(\Leftrightarrow3x-\dfrac{\pi}{4}=k\pi\)

\(\Leftrightarrow3x=\dfrac{\pi}{4}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\)

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NL
13 tháng 6 2021 lúc 0:22

18.

\(\Leftrightarrow3x-\dfrac{\pi}{4}=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow3x=\dfrac{3\pi}{4}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{3}\)

19.

\(\Leftrightarrow\left[{}\begin{matrix}2sin\left(4x-\dfrac{\pi}{12}\right)-1=0\\cos\left(5x-\dfrac{\pi}{3}\right)-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(4x-\dfrac{\pi}{12}\right)=\dfrac{1}{2}\\cos\left(5x-\dfrac{\pi}{3}\right)=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{12}=\dfrac{\pi}{6}+k2\pi\\4x-\dfrac{\pi}{12}=\dfrac{5\pi}{6}+k2\pi\\5x-\dfrac{\pi}{3}=k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{2}\\x=\dfrac{11\pi}{48}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{15}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

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NL
13 tháng 6 2021 lúc 0:26

20.

ĐKXĐ: \(cos\left(4x-\dfrac{\pi}{12}\right)\ne0\Leftrightarrow x\ne\dfrac{7\pi}{48}+\dfrac{k\pi}{4}\)

Pt tương đương:

\(\left[{}\begin{matrix}\sqrt{3}tan\left(4x-\dfrac{\pi}{12}\right)-1=0\\\sqrt{2}cos\left(2x-\dfrac{\pi}{6}\right)-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}tan\left(4x-\dfrac{\pi}{12}\right)=\dfrac{1}{\sqrt{3}}\\cos\left(2x-\dfrac{\pi}{6}\right)=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{12}=\dfrac{\pi}{6}+k\pi\\2x-\dfrac{\pi}{6}=\dfrac{\pi}{4}+k2\pi\\2x-\dfrac{\pi}{6}=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{2}\\x=\dfrac{5\pi}{24}+k\pi\\x=-\dfrac{\pi}{24}+k\pi\end{matrix}\right.\)

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NL
13 tháng 6 2021 lúc 0:27

21.

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(4x-\dfrac{\pi}{12}\right)=0\\cos\left(5x-\dfrac{\pi}{3}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{12}=k\pi\\5x-\dfrac{\pi}{3}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{12}+k\pi\\5x=\dfrac{5\pi}{6}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{48}+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{6}+\dfrac{k\pi}{5}\end{matrix}\right.\)

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NL
13 tháng 6 2021 lúc 0:29

22.

\(\Leftrightarrow\left[{}\begin{matrix}tan\left(4x-\dfrac{\pi}{12}\right)=0\\cot\left(2x-\dfrac{\pi}{6}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{12}=k\pi\\2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{12}+k\pi\\2x=\dfrac{2\pi}{3}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{48}+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}\end{matrix}\right.\)

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