sinx+sin2x+sin3x+sin4x=0

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5 tháng 10 2021

\(\Leftrightarrow\sin x+2\sin x\cos x+3\sin x-4\sin^3x+2\sin2x\cos2x=0\)

\(\Leftrightarrow4\sin x-4\sin^3x+2\sin x\cos x+4\sin x\cos x\left(1-2\sin^2x\right)=0\)

\(\Leftrightarrow4\sin x\left(1-\sin^2x\right)+2\sin x\cos x+4\sin x\cos x-8\sin^3x\cos x=0\)

\(\Leftrightarrow4\sin x\cos^2x+6\sin x\cos x-8\sin^3x\cos x=0\)

\(\Leftrightarrow2\sin x\cos x\left(2\cos x+3+4\sin^2x\right)=0\)

\(\Leftrightarrow\sin2x\left(2\cos x+3+4-4\cos^2x\right)=0\)

Đến đây là phương trình tích. Trong ngoặc đơn là PT bậc 2 đối với cosx. Bạn tự làm nốt nhé

NV
13 tháng 7 2020

\(sin3x-sinx+sin2x=0\)

\(\Leftrightarrow2cos2x.sinx+2sinx.cosx=0\)

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

\(\Leftrightarrow2sinx.cos\frac{3x}{2}.cos\frac{x}{2}=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=0\\cos\frac{x}{2}=0\\cos\frac{3x}{2}=0\end{matrix}\right.\)

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

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

NV
13 tháng 7 2020

\(cosx+cos3x+cos2x+cos4x=0\)

\(\Leftrightarrow2cos2x.cosx+2cos3x.cosx=0\)

\(\Leftrightarrow cosx\left(cos2x+cos3x\right)=0\)

\(\Leftrightarrow2cosx.cos\frac{5x}{2}.cos\frac{x}{2}=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cos\frac{x}{2}=0\\cos\frac{5x}{2}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\\frac{x}{2}=\frac{\pi}{2}+k\pi\\\frac{5x}{2}=\frac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\frac{\pi}{5}+\frac{k2\pi}{5}\end{matrix}\right.\)

NV
21 tháng 8 2020

\(\Leftrightarrow sin4x\left(sin5x+sin3x\right)-sin2x.sinx=0\)

\(\Leftrightarrow2sin^24x.cosx-2sin^2x.cosx=0\)

\(\Leftrightarrow cosx\left(2sin^24x-2sin^2x\right)=0\)

\(\Leftrightarrow cosx\left(1-cos8x-1+cos2x\right)=0\)

\(\Leftrightarrow cosx\left(cos2x-cos8x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos8x=cos2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\8x=2x+k2\pi\\8x=-2x+k2\pi\end{matrix}\right.\)

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

1 tháng 3



\(sin ⁡ 5 x - sin ⁡ 4 x + sin ⁡ 3 x \left(\right. 1 \left.\right)\)

\(\Leftrightarrow \left(\right. sin ⁡ 3 x + sin ⁡ 5 x \left.\right) - sin ⁡ 4 x = 0\)

\(\Leftrightarrow 2 sin ⁡ 4 x . cos ⁡ x - sin ⁡ 4 x = 0\)

\(\Leftrightarrow sin ⁡ 4 x \left(\right. 2 cos ⁡ x - 1 \left.\right) = 0\)

\(\Leftrightarrow \left[\right. sin ⁡ 4 x = 0 \\ 2 cos ⁡ x - 1 = 0 \Leftrightarrow \left[\right. x = \frac{k \pi}{4} \\ x = \frac{\pi}{3} + k 2 \pi \\ x = - \frac{\pi}{3} + k 2 \pi\)\(\left(\right. k \in \mathbb{Z} \left.\right)\)

Vậy các nghiệm của phương trình là \(x = \frac{k \pi}{4}\)\(x = \frac{\pi}{3} + k 2 \pi\) và  \(x = - \frac{\pi}{3} + k 2 \pi\) \(\left(\right. k \in \mathbb{Z} \left.\right)\).

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/

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

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

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

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

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

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

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

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

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

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

NV
15 tháng 7 2020

b/

\(\Leftrightarrow sin3x-sinx-sin3x=1\)

\(\Leftrightarrow sinx=-1\)

\(\Rightarrow x=-\frac{\pi}{2}+k2\pi\)

NV
15 tháng 7 2020

a/ \(\Leftrightarrow sin3x+sinx-sin2x=0\)

\(\Leftrightarrow2sin2x.cosx-sin2x=0\)

\(\Leftrightarrow sin2x\left(2cosx-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=0\\2cosx-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cosx=\frac{1}{2}\end{matrix}\right.\)

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

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

19 tháng 8 2019

Có b nào gipus mk với cần gấp gấp :)

29 tháng 7 2019
https://i.imgur.com/9qSBKHl.jpg
29 tháng 7 2019
https://i.imgur.com/zw6cbvs.jpg
NV
25 tháng 10 2020

1.

\(\Leftrightarrow4sinx.cosx+3\left(sinx-cosx\right)=0\)

Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=1-t^2\end{matrix}\right.\)

Pt trở thành:

\(2\left(1-t^2\right)+3t=0\)

\(\Leftrightarrow-2t^2+3t+2=0\Rightarrow\left[{}\begin{matrix}t=2\left(l\right)\\t=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow sinx-cosx=-\frac{1}{2}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\\x=\frac{5\pi}{4}-arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

NV
25 tháng 10 2020

2.

Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sin2x=2sinx.cosx=1-t^2\end{matrix}\right.\)

Pt trở thành:

\(1-t^2-4t=4\)

\(\Leftrightarrow t^2+4t+3=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sinx-cosx=-1\)

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

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

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

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