Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ ĐKXĐ:
\(sin\left(\frac{\pi}{2}.sinx\right)\ne0\Rightarrow\frac{\pi}{2}.sinx\ne k\pi\)
\(\Rightarrow sinx\ne2k\)
Mà \(-1\le sinx\le1\Rightarrow sinx\ne0\Rightarrow x\ne k\pi\)
b/
\(sinx-1\ge0\Leftrightarrow sinx\ge1\Rightarrow sinx=1\)
\(\Rightarrow x=\frac{\pi}{2}+k2\pi\)
c/
\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\cos2x\ne0\end{matrix}\right.\) \(\Rightarrow sin4x\ne0\)
\(\Rightarrow x\ne\frac{k\pi}{4}\)
d/
\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\sinx+cotx\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ne0\\sin^2x+cosx\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x\ne k\pi\\-cos^2x+cosx+1\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\cosx\ne\frac{1-\sqrt{5}}{2}\\\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\x\ne\pm arccos\left(\frac{1-\sqrt{5}}{2}\right)+k2\pi\end{matrix}\right.\)
e/
\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne1\end{matrix}\right.\) \(\Leftrightarrow sinx\ne0\Rightarrow x\ne k\pi\)
a, \(sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2cos^2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2\cdot\left[1+cos2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right]=0\)
\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-1-cos\left(\dfrac{\pi}{2}-x\right)=0\)
\(\Leftrightarrow sin\dfrac{s}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x-sinx=0\)
\(\Leftrightarrow sinx\cdot\left(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\text{ (1) }\\sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\text{ (2) }\end{matrix}\right.\)
(1) : \(sinx=0\Leftrightarrow x=k\pi\left(k\in Z\right)\)
(2) : \(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-cos\dfrac{x}{2}\cdot2sin\dfrac{x}{2}\cdot cos\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot cos^2\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot\left(1-sin^2\dfrac{x}{2}\right)-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}+2sin^3\dfrac{x}{2}-1=0\)
\(\Leftrightarrow2sin^3\dfrac{x}{2}-sin\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}=1\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\pi+k4\pi\left(k\in Z\right)\)
b, \(tanx-3cotx=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)
\(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{3cos}{sinx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)
\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{sinx-cosx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)
\(\Leftrightarrow sin^2x-3cos^2x=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)
\(\Leftrightarrow\left(sinx-\sqrt{3}\cdot cosx\right)\cdot\left(sinx+\sqrt{3}\cdot cosx\right)=4\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)
\(\Leftrightarrow\left(sinx+\sqrt{3}\cdot cosx\right)\cdot\left[\left(sinx-\sqrt{3}\cdot cosx\right)-4sinx\cdot cosx\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}\cdot cosx=0\text{ (1) }\\sinx-\sqrt{3}\cdot cosx-4sinx\cdot cosx=0\text{ (2) }\end{matrix}\right.\)
(1) : \(sinx+\sqrt{3}\cdot cosx=0\)
\(\Leftrightarrow\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=0\)
\(\Leftrightarrow cos\dfrac{\pi}{3}\cdot sinx+sin\dfrac{\pi}{3}\cdot cosx=0\)
\(\Leftrightarrow sin\cdot\left(x+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\Leftrightarrow x=\dfrac{-\pi}{3}+k\pi\left(k\in Z\right)\)
(2) : \(sinx-\sqrt{3}cosx-4sinx\cdot cosx=0\)
\(\Leftrightarrow sinx-\sqrt{3}cos=2sin2x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cos2=sin2x\)
\(\Leftrightarrow cos\dfrac{\pi}{3}-sinx-sin\dfrac{\pi}{3}\cdot cosx=sin2x\)
\(\Leftrightarrow sin\cdot\left(x-\dfrac{\pi}{3}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=2x+k2\pi\\x-\dfrac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\left(k\in Z\right)\end{matrix}\right.\)
36.
\(sin^2x-cos^2x\ne0\Leftrightarrow cos2x\ne0\)
\(\Leftrightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
37.
\(cos3x\ne cosx\Leftrightarrow\left\{{}\begin{matrix}3x\ne x+k2\pi\\3x\ne-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\frac{k\pi}{2}\end{matrix}\right.\) \(\Leftrightarrow x\ne\frac{k\pi}{2}\)
38.
\(\left\{{}\begin{matrix}x\ge0\\sin\pi x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\pi x\ne k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne k\end{matrix}\right.\)
39.
\(\left\{{}\begin{matrix}cos\left(x-\frac{\pi}{3}\right)\ne0\\tan\left(x-\frac{\pi}{3}\right)\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-\frac{\pi}{3}\ne\frac{\pi}{2}+k\pi\\x-\frac{\pi}{3}\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{5\pi}{6}+k\pi\\x\ne-\frac{\pi}{12}+k\pi\end{matrix}\right.\)
33.
\(\left\{{}\begin{matrix}cosx\ne0\\cos\frac{x}{2}\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\x\ne\pi+k2\pi\end{matrix}\right.\)
34.
\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\cotx\ne1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ne0\\cotx\ne1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\x\ne\frac{\pi}{4}+k\pi\end{matrix}\right.\)
35.
\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne1\end{matrix}\right.\) \(\Leftrightarrow sinx\ne0\)
\(\Leftrightarrow x\ne k\pi\)
3.3 d)
\(\sin8x-\cos6x=\sqrt{3}\left(\sin6x+\cos8x\right)\\ \Leftrightarrow\sin8x-\sqrt{3}\cos8x=\sqrt{3}\sin6x+\cos6x\\ \Leftrightarrow\sin\left(8x-\dfrac{\pi}{3}\right)=\sin\left(6x+\dfrac{\pi}{6}\right)\\ \Leftrightarrow\left[{}\begin{matrix}8x-\dfrac{\pi}{3}=6x+\dfrac{\pi}{6}+k2\pi\\8x-\dfrac{\pi}{3}=\pi-\left(6x+\dfrac{\pi}{6}\right)+k2\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k\dfrac{\pi}{7}\end{matrix}\right.\)
3.4 a)
\(2sin\left(x+\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(\dfrac{\pi}{2}-x-\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(-x+\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(x-\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \)
Chia hai vế cho \(\sqrt{2^2+4^2}=2\sqrt{5}\)
Ta được:
\(\dfrac{1}{\sqrt{5}}cos\left(x-\dfrac{\pi}{4}\right)+\dfrac{2}{\sqrt{5}}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3}{4}\\ \)
Gọi \(\alpha\) là góc có \(cos\alpha=\dfrac{1}{\sqrt{5}}\)và \(sin\alpha=\dfrac{2}{\sqrt{5}}\)
Phương trình tương đương:
\(cos\left(x-\dfrac{\pi}{4}-\alpha\right)=\dfrac{3}{4}\\ \Leftrightarrow x=\pm arscos\left(\dfrac{3}{4}\right)+\dfrac{\pi}{4}+\alpha+k2\pi\)
a: -pi/2<a<0
=>sin a<0
=>sin a=-1/căn 5
tan a=-1/2
cot a=-2
b: pi/2<x<pi
=>cosx<0
=>cosx=-4/5
=>tan x=-3/4
cot x=-4/3
c: -pi<x<-pi/2
=>cosx<0 và sin x<0
1+tan^2x=1/cos^2x
=>1/cos^2x=1+16/25=41/25
=>cosx=-5/căn 41
sin x=-6/căn 41
cot x=5/4
g: 180 độ<x<270 độ
=>cosx <0
=>cosx=-4/5
tan x=3/4
cot x=4/3