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NV
20 tháng 9 2021

\(\Leftrightarrow cos3x=cos\left[\dfrac{\pi}{2}-\left(x-\dfrac{\pi}{4}\right)\right]\)

\(\Leftrightarrow cos3x=cos\left(\dfrac{3\pi}{4}-x\right)\)

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

\(\Rightarrow x=...\)

20 tháng 9 2021

Cảm ơn ạ

7 tháng 6 2019

Bạn tham khảo thử nhé

NV
18 tháng 9 2021

1.

\(sin\left(4x-10^0\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(4x-10^0\right)=sin45^0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-10^0=45^0+k360^0\\4x-10^0=135^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=55^0+k360^0\\4x=145^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=13,75^0+k90^0\\x=36,25^0+k90^0\end{matrix}\right.\) (\(k\in Z\))

NV
18 tháng 9 2021

2.

Đề không đúng

3.

ĐKXĐ: \(\left\{{}\begin{matrix}cos2x\ne0\\cosx\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(tan2x=tanx\)

\(\Rightarrow2x=x+k\pi\)

\(\Rightarrow x=k\pi\)

4.

\(cot\left(x+\dfrac{\pi}{5}\right)=-1\)

\(\Leftrightarrow x+\dfrac{\pi}{5}=-\dfrac{\pi}{4}+k\pi\)

\(\Leftrightarrow x=-\dfrac{9\pi}{20}+k\pi\) (\(k\in Z\))

17 tháng 10 2021

\(sin\left(2x+\dfrac{\pi}{3}\right)+cos3x=0\)

\(\Leftrightarrow cos\left(\dfrac{\pi}{6}-2x\right)+cos3x=0\)

\(\Leftrightarrow2cos\left(\dfrac{\pi}{12}+\dfrac{x}{2}\right).cos\left(\dfrac{\pi}{12}-\dfrac{5x}{2}\right)=0\)

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

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

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

17 tháng 10 2021

Ta có: \(sin\left(2x+\dfrac{\pi}{3}\right)=-cos3x=cos\left(\pi-3x\right)=sin\left(\dfrac{\pi}{2}-\left(\pi-3x\right)\right)=sin\left(3x-\dfrac{1}{2}\right)\)

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=3x-\dfrac{1}{2}+k2\pi\\2x+\dfrac{\pi}{3}=\pi-3x+\dfrac{1}{2}+k2\pi\end{matrix}\right.\) Bạn tự tìm x được.

1 tháng 10 2021

\(cos3x=sin\left(x+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow cos3x=cos\left(\dfrac{\pi}{4}-x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{8}+k\pi\end{matrix}\right.\)

30 tháng 9 2021

Pt\(\Rightarrow cos3x=cos[\dfrac{\pi}{2}-(x+\dfrac{\pi}{4})]\)

   \(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{2}-(x+\dfrac{\pi}{4})+k2\pi\\3x=-\dfrac{\pi}{2}+\left(x+\dfrac{\pi}{4}\right)+k2\pi\end{matrix}\right.\)

   \(\Rightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{4}+k2\pi\\2x=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+k2\pi\\x=-\dfrac{\pi}{8}+k2\pi\end{matrix}\right.\)(k\(\in\)Z)

4 tháng 3 2018

a) √2 cos(x - π/4)

= √2.(cosx.cos π/4 + sinx.sin π/4)

= √2.(√2/2.cosx + √2/2.sinx)

= √2.√2/2.cosx + √2.√2/2.sinx

= cosx + sinx (đpcm)

b) √2.sin(x - π/4)

= √2.(sinx.cos π/4 - sin π/4.cosx )

= √2.(√2/2.sinx - √2/2.cosx )

= √2.√2/2.sinx - √2.√2/2.cosx

= sinx – cosx (đpcm).