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29 tháng 7 2020

\(\text{1) }3sinx-4cosx=1\\ \Leftrightarrow cos^2x+\left(\frac{4cosx+1}{3}\right)^2=1\\ \Leftrightarrow cosx=\frac{-4\pm6\sqrt{6}}{25}\\ \\ \Leftrightarrow x=arccos\left(\frac{-4\pm6\sqrt{6}}{25}\right)+k2\pi\)

\(2\text{) }\sqrt{3}sinx-cosx=1\\ \Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{1}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sinx-sin\frac{\pi}{6}\cdot cosx=\frac{1}{2}\\ \Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=sin\frac{\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=\frac{\pi}{6}+a2\pi\\x-\frac{\pi}{6}=\frac{5\pi}{6}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+a2\pi\\x=\pi+b2\pi\end{matrix}\right.\)

\(3\text{) }\sqrt{3}cosx+sinx=-2\\ \Leftrightarrow\frac{\sqrt{3}}{2}cosx+\frac{1}{2}sinx=-1\\ \Leftrightarrow sin\frac{\pi}{3}\cdot cosx+cos\frac{\pi}{3}\cdot sinx=-1\\ \Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=-1=sin\frac{3\pi}{2}\\ \\ \Leftrightarrow x+\frac{\pi}{3}=\frac{3\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{7\pi}{6}+k2\pi\)

\(4\text{) }cos4x-sin4x=1\\ \Leftrightarrow cos^24x+\left(cos4x-1\right)^2=1\\ \\ \Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+a\pi\\4x=b2\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{a\pi}{4}\\x=\frac{b\pi}{2}\end{matrix}\right.\)

29 tháng 7 2020

\(5\text{) }\sqrt{3}cos4x+sin4x-2cos3x=0\\ \Leftrightarrow\frac{\sqrt{3}}{2}cos4x+\frac{1}{2}sin4x=cos3x\\ \Leftrightarrow cos\frac{\pi}{3}\cdot cos4x+sin\frac{\pi}{3}\cdot sin4x=cos3x\\ \Leftrightarrow cos\left(4x-\frac{\pi}{3}\right)=cos3x\\ \Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{3}=3x+a2\pi\\4x-\frac{\pi}{3}=-3x+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+a2\pi\\x=\frac{\pi}{21}+\frac{b2\pi}{7}\end{matrix}\right.\\ \Leftrightarrow x=\frac{\pi}{21}+\frac{k2\pi}{7}\)

\(6\text{) }cos^2x=3sin2x+3\\ \Leftrightarrow\frac{cos2x+1}{2}=3sin2x+3\)

Giải tương tự vd 1 và 4

7) Giải tương tự vd 1 và 4

20 tháng 8 2021

1.

\(2sin\left(x+10^o\right)-\sqrt{12}cos\left(x+10^o\right)=3\)

\(\Leftrightarrow\dfrac{1}{2}sin\left(x+10^o\right)-\dfrac{\sqrt{3}}{2}cos\left(x+10^o\right)=\dfrac{3}{4}\)

\(\Leftrightarrow sin\left(x+50^o\right)=\dfrac{3}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+50^o=arcsin\left(\dfrac{3}{4}\right)+k360^o\\x+50^o=180^o-arcsin\left(\dfrac{3}{4}\right)+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-50^o+arcsin\left(\dfrac{3}{4}\right)+k360^o\\x=130^o-arcsin\left(\dfrac{3}{4}\right)+k360^o\end{matrix}\right.\)

20 tháng 8 2021

2.

\(\sqrt{3}sin4x-cos4x=\sqrt{3}\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin4x-\dfrac{1}{2}cos4x=\dfrac{\sqrt{3}}{2}\)

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

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

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

NV
30 tháng 8 2020

c/

\(\Leftrightarrow cos3x-\sqrt{3}sin3x=\sqrt{3}cos2x-sin2x\)

\(\Leftrightarrow\frac{1}{2}cos3x-\frac{\sqrt{3}}{2}sin3x=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)

\(\Leftrightarrow cos\left(3x+\frac{\pi}{3}\right)=cos\left(2x+\frac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{3}=2x+\frac{\pi}{6}+k2\pi\\3x+\frac{\pi}{3}=-2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=-\frac{\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)

NV
30 tháng 8 2020

b/

\(\Leftrightarrow cosx-\sqrt{3}sinx=sin2x-\sqrt{3}cos2x\)

\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)

\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=sin\left(2x-\frac{\pi}{3}\right)\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=sin\left(\frac{\pi}{6}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{5\pi}{6}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

NV
16 tháng 7 2021

a.

\(\Leftrightarrow\dfrac{\sqrt{2}}{2}sin4x+\dfrac{\sqrt{2}}{2}cos4x=\dfrac{\sqrt{6}}{2}\)

\(\Leftrightarrow cos4x.cos\left(\dfrac{\pi}{4}\right)+sin4x.sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{6}}{2}\)

\(\Leftrightarrow cos\left(4x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{6}}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{1}{4}arccos\left(\dfrac{\sqrt{6}}{2}\right)+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{16}-\dfrac{1}{4}arccos\left(\dfrac{\sqrt{6}}{2}\right)+\dfrac{k\pi}{4}\end{matrix}\right.\)

NV
16 tháng 7 2021

b.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx=\dfrac{\sqrt{3}}{6}\)

\(\Leftrightarrow cosx.cos\left(\dfrac{\pi}{3}\right)+sinx.sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{6}\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{6}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+arccos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\\x=\dfrac{\pi}{3}-arrcos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\end{matrix}\right.\)

NV
20 tháng 9 2019

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin4x-\frac{1}{2}cos4x=\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx\)

\(\Leftrightarrow sin4x.cos\frac{\pi}{6}-cos4x.sin\frac{\pi}{6}=sinx.cos\frac{\pi}{3}-cosx.sin\frac{\pi}{3}\)

\(\Leftrightarrow sin\left(4x-\frac{\pi}{6}\right)=sin\left(x-\frac{\pi}{3}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{18}+\frac{k2\pi}{3}\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)

1 tháng 6 2021

1.

\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)

\(\Leftrightarrow sinx-\sqrt{3}cosx=\sqrt{2}cos3x+\sqrt{2}sin3x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{\sqrt{2}}cos3x+\dfrac{1}{\sqrt{2}}sin3x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7\pi}{24}-k\pi\\x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{7\pi}{24}-k\pi;x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\)

1 tháng 6 2021

2.

\(sinx-\sqrt{3}cosx=2sin5\text{​​}x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin5x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin5x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)

19 tháng 9 2017

hộ vs ae ơi