Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a.
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(sinx+cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=-1\\2cosx-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\cosx=\frac{3}{2}\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
\(\Leftrightarrow1+sinx+cosx+2sinx.cosx+2cos^2x-1=0\)
\(\Leftrightarrow sinx\left(2cosx+1\right)+cosx\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
c/
Đặt \(3cosx-4sinx-6=t\)
Pt trở thành:
\(t^2+2=-3t\Leftrightarrow t^2+3t+2=0\)
\(\Rightarrow\left[{}\begin{matrix}t=-1\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3cosx-4sinx-6=-1\\3cosx-4sinx-6=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3cosx-4sinx=5\\3cosx-4sinx=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx.\frac{3}{5}-sinx.\frac{4}{5}=1\\cosx.\frac{3}{5}-sinx.\frac{4}{5}=\frac{4}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x+a\right)=1\\cosx\left(x+a\right)=\frac{4}{5}\end{matrix}\right.\) (với góc \(a\in\left[0;\pi\right]\) sao cho \(cosa=\frac{3}{5}\))
\(\Leftrightarrow\left[{}\begin{matrix}x+a=k2\pi\\x+a=\pm\left(\frac{\pi}{2}-a\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-a+k2\pi\\x=-a\pm\left(\frac{\pi}{2}-a\right)+k2\pi\end{matrix}\right.\)
a/
\(\Leftrightarrow cosx.\frac{1}{2}-\frac{\sqrt{3}}{2}sinx=cos\left(\frac{\pi}{3}-x\right)\)
\(\Leftrightarrow cosx.cos\left(\frac{\pi}{3}\right)-sinx.sin\left(\frac{\pi}{3}\right)=cos\left(\frac{\pi}{3}-x\right)\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{3}-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{3}-x+k2\pi\\x+\frac{\pi}{3}=-\frac{\pi}{3}+x+k2\pi\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=k\pi\)
b/
\(\Leftrightarrow\sqrt{2}sin\left(5x+\frac{\pi}{4}\right)=\sqrt{2}cos13x\)
\(\Leftrightarrow cos\left(\frac{\pi}{4}-5x\right)=cos13x\)
\(\Leftrightarrow\left[{}\begin{matrix}13x=\frac{\pi}{4}-5x+k2\pi\\13x=-\frac{\pi}{4}+5x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{72}+\frac{k\pi}{9}\\x=-\frac{\pi}{32}+\frac{k\pi}{4}\end{matrix}\right.\)
2.1
a.
\(\Leftrightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)
b.
\(cosx-\sqrt{3}sinx=1\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
1) a) cos7x - √3 sin7x = -√2 (a = 1; b = -√3; c = -√2)
=> a^2 + b^2 =4 > c^2 = 2
Chia 2 vế pt (*) cho \(\sqrt{a^2+b^2}=2\) ta đc:
<=> 1/2cos7x - √3/2 sin7x = -√2/2
<=> sin(π/6)cos7x - cos(π/6)sin7x = sin(-π/4)
<=> sin(π/6 - 7x) = sin(-π/4)
<=> π/6 - 7x = -π/4 + k2π
hoặc (k∈Z)
π/6 - 7x = π + π/4 + k2π
<=> x = 5π/84 + k2π/7
hoặc (k∈Z)
x = -13π/84 + k2π/7
1) b) Ta có:
* 2π/5 < x < 6π/7
<=> 2π/5 < 5π/84 + k2π/7 < 6π/7
<=> 143π/420 < k2π/7 < 67π/84
<=> 143/120 < k < 67/24
=> k ϵ {2}
=> x = 53π/84
* 2π/5 < x < 6π/7
<=> 2π/5 < -13π/84 + k2π/7 < 6π/7
<=> 233/120 < k < 85/24
=> k ϵ {2; 3}
=> x = 5π/12 ; x = 59π/84
Vậy có tất cả 3 nghiệm thỏa mãn (2π/5;6π/7) là x = 53π/84; x = 5π/12 ; x = 59π/84.
7.
\(\Leftrightarrow\left[{}\begin{matrix}2x-40^0=60^0+k360^0\\2x-40^0=120^0+n360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=50^0+k180^0\\x=80^0+n180^0\end{matrix}\right.\)
Do \(-180^0\le x\le180^0\Rightarrow\left\{{}\begin{matrix}-180^0\le50^0+k180^0\le180^0\\-180^0\le80^0+n180^0\le180^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-\frac{23}{18}\le k\le\frac{13}{18}\\-\frac{13}{9}\le n\le\frac{5}{9}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}k=\left\{-1;0\right\}\\n=\left\{-1;0\right\}\end{matrix}\right.\)
\(\Rightarrow x=\left\{-130^0;50^0;-100^0;80^0\right\}\)
8.
\(\Leftrightarrow sinx=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k2\pi\\x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
5.
\(\Leftrightarrow\frac{\sqrt{2}}{2}sin2x+\frac{\sqrt{2}}{2}cos2x=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin2x.sin\frac{\pi}{4}+cos2x.cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\2x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
6.
\(\Leftrightarrow2sin2x=-1\)
\(\Leftrightarrow sin2x=-\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-\frac{\pi}{6}+k2\pi\\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)
a.\(\dfrac{sin2x+cosx-\sqrt{3}\left(cos2x+sinx\right)}{2sin2x-\sqrt{3}}=1\left(1\right)\)
ĐKXĐ: sin2x≠\(\dfrac{\sqrt{3}}{2}\)
(1) ⇔ sin2x + cosx - \(\sqrt{3}\) ( cos2x + sinx) = 2sin2x - \(\sqrt{3}\)
⇔cosx - \(\sqrt{3}\) sinx = \(\sqrt{3}\) cos2x + sin2x +\(\sqrt{3}\)
⇔\(\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}\)
⇔\(sin\left(\dfrac{\Pi}{6}-x\right)=sin\left(2x+\dfrac{\Pi}{3}\right)-sin\dfrac{\Pi}{3}\)
⇔\(sin\left(\dfrac{\Pi}{6}-x\right)=2cos\left(x+\dfrac{\Pi}{3}\right)sinx\)
⇔\(sin\left(\dfrac{\Pi}{6}-x\right)=2sin\left(\dfrac{\Pi}{6}-x\right)sinx\)
⇔\(sin\left(\dfrac{\Pi}{6}-x\right)\left(2sinx-1\right)=0\)
Đến đây tự giải tiếp nha nhớ đối chiếu đk.
b.\(\left(2cosx-1\right)cotx=\dfrac{3}{sinx}+\dfrac{2sinx}{cosx-1}\left(1\right)\)
ĐKXĐ: sinx≠0 và cosx≠1
(1)⇔\(\left(2cosx-1\right)\dfrac{cosx}{sinx}=\dfrac{3}{sinx}+\dfrac{2sinx}{cosx-1}\)
⇔cosx(2cosx-1)(cosx-1) = 3(cosx-1) + 2sin2x
⇔2cos3x - cos2x - 2cosx +1 = 0
⇔ (cosx-1)(cosx+1)(2cosx-1)=0
Giải phương trình: \(\frac{cosx\left(cosx+2sinx\right)+3sinx\left(sĩn+\sqrt{2}\right)}{2sinx-1}\)= 1
Cái chỗ biến đổi tương đương cuối cùng bạn làm rõ chút dc ko???
Ném đoạn \(2sin^2x+\left(3\sqrt{2}-2\right)sinx+1\) vào casio mà bấm pt bậc 2 thôi, nó sẽ tách ra biểu thức như cái cuối cùng
Hoặc là tách thế này:
\(2sin^2x+\left(3\sqrt{2}-2\right)sinx+1\)
\(=2\left[sin^2x-2.\frac{2-3\sqrt{2}}{4}sinx+\left(\frac{2-3\sqrt{2}}{4}\right)^2-\left(\frac{2-3\sqrt{2}}{4}\right)^2\right]+1\)
\(=2\left(sinx-\frac{2-3\sqrt{2}}{4}\right)^2-2\left(\frac{2-3\sqrt{2}}{4}\right)^2+1\)
\(=2\left(sin^2x-\frac{2-3\sqrt{2}}{4}\right)^2+\frac{6\sqrt{2}-7}{4}\)
Với lưu ý \(\frac{6\sqrt{2}-7}{4}>0\) nên biểu thức luôn dương
d/
\(\Leftrightarrow\frac{2}{\sqrt{29}}sinx-\frac{5}{\sqrt{29}}cosx=\frac{5}{\sqrt{29}}\)
Đặt \(cosa=\frac{2}{\sqrt{29}}\) với \(0< a< \pi\)
\(\Rightarrow sinx.cosa-cosx.sina=sina\)
\(\Leftrightarrow sin\left(x-a\right)=sina\)
\(\Rightarrow\left[{}\begin{matrix}x-a=a+k2\pi\\x-a=\pi-a+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2a+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\frac{\sqrt{3}}{\sqrt{19}}cosx+\frac{4}{\sqrt{19}}sinx=\frac{\sqrt{3}}{\sqrt{19}}\)
Đặt \(cosa=\frac{\sqrt{3}}{\sqrt{19}}\) với \(0< a< \pi\)
\(\Rightarrow cosx.cosa+sinx.sina=cosa\)
\(\Leftrightarrow cos\left(x-a\right)=cosa\)
\(\Rightarrow\left[{}\begin{matrix}x-a=a+k2\pi\\x-a=-a+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2a+k2\pi\\x=k2\pi\end{matrix}\right.\)