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15 tháng 7 2020

\(\text{c) }sin3x-\sqrt{3}cos3x=2cos5x\\ \Leftrightarrow\frac{1}{2}sin3x-\frac{\sqrt{3}}{2}cos3x=cos5x\\ \Leftrightarrow sin\frac{\pi}{6}\cdot sin3x-cos\frac{\pi}{6}\cdot cos3x=cos5x\\ \Leftrightarrow cos\left(3x+\frac{\pi}{6}\right)=-cos5x\\ \Leftrightarrow cos\left(3x+\frac{\pi}{6}\right)=cos\left(\pi-5x\right)\\ \Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{6}=\pi-5x+m2\pi\\3x+\frac{\pi}{6}=5x-\pi+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{48}+\frac{m\pi}{4}\\x=\frac{7\pi}{12}-n\pi\end{matrix}\right.\)

\(d\text{) }sinx\left(sinx+2cosx\right)=2\\ \Leftrightarrow cos^2x+\left(sinx-cosx\right)^2=0\\ \Leftrightarrow cosx=sinx=0\left(VN\right)\)

\(e\text{) }\sqrt{3}\left(sin2x+cos7x\right)=sin7x-cos2x\\ \Leftrightarrow\sqrt{3}sin2x+cos2x=sin7x-\sqrt{3}cos7x\\ \Leftrightarrow sin2x\cdot\frac{\sqrt{3}}{2}+cos2x\cdot\frac{1}{2}=sin7x\cdot\frac{1}{2}-cos7x\cdot\frac{\sqrt{3}}{2}\\ \Leftrightarrow sin2x\cdot cos\frac{\pi}{3}+cos2x\cdot sin\frac{\pi}{3}=sin7x\cdot cos\frac{\pi}{3}-cos7x\cdot sin\frac{\pi}{3}\\ \Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=sin\left(7x-\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=7x-\frac{\pi}{3}+m2\pi\\2x-\frac{\pi}{3}=\frac{4\pi}{3}-7x+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-m2\pi}{5}\\x=\frac{5\pi}{27}+\frac{n2\pi}{9}\end{matrix}\right.\)

15 tháng 7 2020

\(\text{a) }\sqrt{3}sin2x-cos2x+1=0\\ \Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x=-\frac{1}{2}\\ \Leftrightarrow cos\frac{\pi}{3}\cdot cos2x-sin\frac{\pi}{3}\cdot sin2x=\frac{1}{2}\\ \Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=cos\frac{\pi}{3}\\ \Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{3}+m2\pi\\2x-\frac{\pi}{3}=-\frac{\pi}{3}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+m\pi\\x=n\pi\end{matrix}\right.\)

\(\text{b) }pt\Leftrightarrow sin4x=\frac{1-4cosx}{3}\\ \Leftrightarrow sin^24x+cos^24x=\left(\frac{1-cos4x}{3}\right)^2+cos^24x=1\\ \Leftrightarrow\left[{}\begin{matrix}cos4x=1\\cos4x=-\frac{4}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cos4x=1\\cos4x=-\frac{4}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{arccos\left(-\frac{4}{5}\right)}{4}+\frac{k\pi}{2}\end{matrix}\right.\)

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/

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

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

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

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

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

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

28 tháng 9 2020

@Nguyễn Việt Lâm giúp em với ạ

NV
29 tháng 9 2020

a/ \(4cos^3x-3cosx-4\left(2cos^2x-1\right)+3cosx-4=0\)

\(\Leftrightarrow4cos^3x-8cos^2x=0\)

\(\Leftrightarrow4cos^2x\left(cosx-2\right)=0\)

\(\Leftrightarrow cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\)

\(0< \frac{\pi}{2}+k\pi< 14\Rightarrow-\frac{1}{2}< k< \frac{14-\frac{\pi}{2}}{\pi}\Rightarrow k=\left\{0;1;2;3\right\}\)

\(\Rightarrow x=\left\{\frac{\pi}{2};\frac{3\pi}{2};\frac{5\pi}{2};\frac{7\pi}{2}\right\}\)

b/ Bạn coi lại đề, cái ngoặc thứ 2 thiếu \(\left(2cos\left(???\right)+cosx\right)\)

c/ Bạn coi lại đề, có 2 số hạng \(cos2x\) xuất hiện ở vế trái, cấp 3 chắc ko ai cho kiểu vậy đâu, nếu đúng thế thì người ta cộng luôn thành \(2cos2x\) cho rồi

b: \(\Leftrightarrow2\cdot\cos2x\cdot\cos x+2\cdot\sin x\cdot\cos2x=\sqrt{2}\cdot\cos2x\)

\(\Leftrightarrow2\cdot\cos2x\left(\sin x+\cos x\right)=\sqrt{2}\cdot\cos2x\)

\(\Leftrightarrow\sqrt{2}\cdot\cos2x\cdot\left[\sqrt{2}\cdot\sqrt{2}\cdot\sin\left(x+\dfrac{\Pi}{4}\right)-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\cos2x=0\\\sin\left(x+\dfrac{\Pi}{4}\right)=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Pi}{2}+k\Pi\\x+\dfrac{\Pi}{4}=\dfrac{\Pi}{6}+k2\Pi\\x+\dfrac{\Pi}{4}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\)

\(\Leftrightarrow x\in\left\{\dfrac{\Pi}{4}+\dfrac{k\Pi}{2};\dfrac{-1}{12}\Pi+k2\Pi;\dfrac{7}{12}\Pi+k2\Pi\right\}\)

c: \(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=2\cdot\cos2x\cdot\cos x+\cos2x\)

\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=\cos2x\left(2\cos x+1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\cos2x=\sin\left(\dfrac{\Pi}{2}-2x\right)\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{4}\\\\x=-\dfrac{2}{3}\Pi+k2\Pi\\x=\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)

29 tháng 9 2020

a.\(\frac{k\Pi}{2}+\frac{\alpha}{2}\)

b.\(\left\{{}\begin{matrix}x=\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\\x=\Pi-\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\end{matrix}\right.\)

NV
6 tháng 10 2020

a.

\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)

\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0=0\)

\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)

\(\Leftrightarrow\left(sinx+cosx+1\right)\left(2cosx-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=-1\\2cosx-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\cosx=\frac{3}{2}\left(vn\right)\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

NV
6 tháng 10 2020

b.

\(\Leftrightarrow1+sinx+cosx+2sinx.cosx+2cos^2x-1=0\)

\(\Leftrightarrow sinx\left(2cosx+1\right)+cosx\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)

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

26 tháng 2 2018

Chọn D

Ta sẽ biến đổi phương trình thành dạng tích

Chú ý: có thể dùng 4 đáp án thay vào phương trình để kiểm tra đâu là nghiệm

1 tháng 6 2021

1.

\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)

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

\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)

\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)

Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)

\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm:

\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)

1 tháng 6 2021

2.

\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)

\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)

\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)