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18 tháng 10 2021

\(3tan\cdot\left(4x-\dfrac{\pi}{3}\right)+\sqrt{3}=0\)

\(\Rightarrow tan\left(4x-\dfrac{\pi}{3}\right)=-\dfrac{1}{\sqrt{3}}=tan\left(-\dfrac{\pi}{6}\right)\)

\(\Rightarrow4x-\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k\pi\)\(\Rightarrow4x=\dfrac{\pi}{6}+k\pi\)

\(\Rightarrow x=\dfrac{\pi}{24}+k\dfrac{\pi}{4}\left(k\in Z\right)\)

6 tháng 8 2021

a, y xác định `<=> 3cos(2x+3) \ne 0`

`<=>cos(2x+3) \ne 0`

`<=>2x+3 \ne π/2+kπ`

`<=>x \ne π/4 -3/2 +k π/2 (k \in ZZ)`

b, y xác định `<=> sin(x/3+π/4) \ne0`

`<=> x/3+π/4 \ne kπ`

`<=> x \ne (-3π)/4+ k3π`

NV
6 tháng 8 2021

ĐKXĐ: 

a.

\(cos\left(2x+3\right)\ne0\)

\(\Leftrightarrow2x+3\ne\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=-\dfrac{3}{2}+\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

b.

\(sin\left(\dfrac{x}{3}+\dfrac{\pi}{4}\right)\ne0\)

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

\(\Leftrightarrow x\ne-\dfrac{3\pi}{4}+k3\pi\)

16 tháng 7 2020

\(\frac{tanx-1}{tanx+1}+cot2x=0\\ \Leftrightarrow cot2x-\frac{1-tanx\cdot tan\frac{\pi}{4}}{tanx+tan\frac{\pi}{4}}=0\\ \Leftrightarrow cot2x-cot\left(x+\frac{\pi}{4}\right)=0\)

NV
16 tháng 7 2020

d/

ĐKXĐ: \(\left\{{}\begin{matrix}sin2x\ne0\\tanx\ne-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\x\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{tanx-1}{tanx+1}+cot2x=0\\3tanx-\sqrt{3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{tanx-1}{tanx+1}-\frac{tan^2x-1}{2tanx}=0\\tanx=\frac{\sqrt{3}}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(tanx-1\right)\left(\frac{1}{tanx+1}-\frac{tanx+1}{2tanx}\right)=0\left(1\right)\\x=\frac{\pi}{6}+k\pi\end{matrix}\right.\)

Xét (1): \(\Leftrightarrow\left[{}\begin{matrix}tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\\\frac{1}{tanx+1}-\frac{tanx+1}{2tanx}=0\left(2\right)\end{matrix}\right.\)

Xét (2)

\(\Leftrightarrow\left(tanx+1\right)^2-2tanx=0\)

\(\Leftrightarrow tan^2x+1=0\left(vn\right)\)

NV
19 tháng 7 2020

c/

\(\Leftrightarrow tan\left(60^0-x\right)=-\frac{1}{\sqrt{3}}\)

\(\Rightarrow60^0-x=-30^0+k180^0\)

\(\Rightarrow x=90^0+k180^0\)

d/

\(\Leftrightarrow tan\left(3x+\frac{2\pi}{5}\right)=-tan\left(\frac{\pi}{5}\right)\)

\(\Leftrightarrow tan\left(3x+\frac{2\pi}{5}\right)=tan\left(-\frac{\pi}{5}\right)\)

\(\Rightarrow3x+\frac{2\pi}{5}=-\frac{\pi}{5}+k\pi\)

\(\Rightarrow x=-\frac{\pi}{5}+\frac{k\pi}{3}\)

NV
19 tháng 7 2020

a/

\(\Leftrightarrow tan2x=-tan40^0\)

\(\Leftrightarrow tan2x=tan\left(-40^0\right)\)

\(\Rightarrow2x=-40^0+k180^0\)

\(\Rightarrow x=-20^0+k90^0\)

b/

\(\Leftrightarrow tan\left(2x-15^0\right)=1\)

\(\Rightarrow2x-15^0=45^0+k180^0\)

\(\Rightarrow x=30^0+k90^0\)

1 tháng 10 2021

=-1

8 tháng 2 2022

a, ĐK: \(x\ne\dfrac{5\pi}{6}+k2\pi;x\ne\dfrac{\pi}{6}+k2\pi\)

\(\dfrac{2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)}{2sinx-1}=-1\)

\(\Leftrightarrow2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)=1-2sinx\)

\(\Leftrightarrow-cos\left(3x-\dfrac{\pi}{2}\right)+\sqrt{3}cos^3x.\dfrac{cos^2x-3sin^2x}{cos^2x}=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(cos^2x-3sin^2x\right)=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(4cos^2x-3\right)=-2sinx\)

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

\(\Leftrightarrow\dfrac{1}{2}sin3x-\dfrac{\sqrt{3}}{2}cos3x-sinx=0\)

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

\(\Leftrightarrow2cos\left(2x-\dfrac{\pi}{6}\right)sin\left(x-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(2x-\dfrac{\pi}{6}\right)=0\\sin\left(x-\dfrac{\pi}{6}\right)=0\end{matrix}\right.\)

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

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

Đối chiếu điều kiện ta được:

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

NV
15 tháng 2 2022

(Giả sử chọn k=-1)

Đặt \(u_n=v_n-1\Rightarrow v_{n+1}-1=\dfrac{5\left(v_n-1\right)+4}{v_n-1+2}=\dfrac{5v_n-1}{v_n+1}\)

\(\Rightarrow v_{n+1}=1+\dfrac{5v_n-1}{v_n+1}=\dfrac{6v_n}{v_n+1}\)

Mục đích chỉ cần biến đổi tới đây, sau đó nghịch đảo 2 vế:

\(\Rightarrow\dfrac{1}{v_{n+1}}=\dfrac{v_n+1}{6v_n}=\dfrac{1}{6v_n}+\dfrac{1}{6}\)

Đặt \(\dfrac{1}{v_n}=x_n\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{1}{v_1}=\dfrac{1}{u_1+1}=\dfrac{1}{6}\\x_{n+1}=\dfrac{1}{6}x_n+\dfrac{1}{6}\end{matrix}\right.\)

Rồi đó, đưa về dãy cơ bản \(\Rightarrow x_{n+1}-\dfrac{1}{5}=\dfrac{1}{6}\left(x_n-\dfrac{1}{5}\right)\)

Đặt \(x_n-\dfrac{1}{5}=y_n\Rightarrow\left\{{}\begin{matrix}y_1=x_1-\dfrac{1}{5}=-\dfrac{1}{30}\\y_{n+1}=\dfrac{1}{6}y_n\end{matrix}\right.\)

\(\Rightarrow y_n=-\dfrac{1}{30}\left(\dfrac{1}{6}\right)^{n-1}\Rightarrow x_n=y_n+\dfrac{1}{5}=-\dfrac{1}{30}.\left(\dfrac{1}{6}\right)^{n-1}+\dfrac{1}{5}\)

\(\Rightarrow v_n=\dfrac{1}{x_n}=...\Rightarrow u_n=v_n-1=\dfrac{1}{x_n}-1=...\)

Cách này là cách cơ bản, có hướng làm cố định để đưa về các dãy quen thuộc

NV
18 tháng 9 2021

\(\Leftrightarrow cos\left(4x+\dfrac{\pi}{3}\right)=-sin\left(x+\dfrac{\pi}{5}\right)\)

\(\Leftrightarrow cos\left(4x+\dfrac{\pi}{3}\right)=cos\left(x+\dfrac{7\pi}{10}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+\dfrac{\pi}{3}=x+\dfrac{7\pi}{10}+k2\pi\\4x+\dfrac{\pi}{3}=-x-\dfrac{7\pi}{10}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{11\pi}{30}+k2\pi\\5x=-\dfrac{31\pi}{30}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11\pi}{90}+\dfrac{k2\pi}{3}\\x=-\dfrac{31\pi}{150}+\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))

12 tháng 6 2021

ĐK: \(x\ne k\pi\)

\(3tan^2\left(x-\dfrac{\pi}{2}\right)=2.\dfrac{1-sinx}{sinx}\)

\(\Leftrightarrow3cot^2x=\dfrac{2}{sinx}-2\)

\(\Leftrightarrow\dfrac{3}{sin^2x}-3=\dfrac{2}{sinx}-2\)

\(\Leftrightarrow\dfrac{3}{sin^2x}-\dfrac{2}{sinx}-1=0\)

\(\Leftrightarrow\left(\dfrac{1}{sinx}-1\right)\left(\dfrac{3}{sinx}+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-3\left(l\right)\end{matrix}\right.\)

\(sinx=1\Leftrightarrow x=\dfrac{\pi}{2}+\text{k}2\pi\left(tm\right)\)

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

NV
12 tháng 6 2021

ĐKXĐ: \(x\ne k\pi\)

\(3cot^2x=2\left(\dfrac{1-sinx}{sinx}\right)\)

\(\Leftrightarrow\dfrac{3cos^2x}{sin^2x}=2\left(\dfrac{1-sinx}{sinx}\right)\)

\(\Leftrightarrow\dfrac{3\left(1-sinx\right)\left(1+sinx\right)}{sin^2x}-2\left(\dfrac{1-sinx}{sinx}\right)=0\)

\(\Leftrightarrow\left(\dfrac{1-sinx}{sinx}\right)\left(\dfrac{3+3sinx}{sinx}-2\right)=0\)

\(\Leftrightarrow\left(\dfrac{1-sinx}{sinx}\right)\left(\dfrac{3+sinx}{sinx}\right)=0\)

\(\Leftrightarrow sinx=1\)

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

NV
26 tháng 9 2019

ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)

\(\frac{3sin^2x}{cos^2x}+\frac{3\left(sinx+cosx\right)}{cos^2x}=1+4\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)

\(\Leftrightarrow\frac{3sin^2x}{cos^2x}+\frac{3\left(sinx+cosx\right)}{cos^2x}=1+4\left(sinx+cosx\right)\)

\(\Leftrightarrow\frac{3-3cos^2x}{cos^2x}-1+\frac{3\left(sinx+cosx\right)}{cos^2x}-4\left(sinx+cosx\right)=0\)

\(\Leftrightarrow\frac{3-4cos^2x}{cos^2x}+\left(sinx+cosx\right)\left(\frac{3-4cos^2x}{cos^2x}\right)=0\)

\(\Leftrightarrow\left(\frac{3-4cos^2x}{cos^2x}\right)\left(sinx+cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3-4cos^2x=0\\sinx+cosx=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=\frac{3}{4}\\\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{3}}{2}\\cosx=\frac{-\sqrt{3}}{2}\\sin\left(x+\frac{\pi}{4}\right)=\frac{-\sqrt{2}}{2}\end{matrix}\right.\) \(\Rightarrow...\)

NV
22 tháng 1

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^22x-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)

\(\Leftrightarrow1-\dfrac{1}{2}\left(\dfrac{1-cos4x}{2}\right)-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)

\(\Leftrightarrow-\dfrac{3}{4}-\dfrac{1}{4}cos4x+\dfrac{1}{2}sin2x=0\)

\(\Leftrightarrow-\dfrac{3}{4}-\dfrac{1}{4}\left(1-2sin^22x\right)+\dfrac{1}{2}sin2x=0\)

\(\Leftrightarrow...\)