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ngại viết quá hihi, mà hơi ngáo tí cái dạng này lm rồi mà cứ quên
bài trước mk bình luận bạn đọc chưa nhỉ
c/
\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{\sqrt{2}}\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}3x-\frac{\pi}{4}=\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{4}=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{7\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{36}+\frac{k2\pi}{3}\end{matrix}\right.\)
d/
\(\Leftrightarrow2sinx.cosx+1-2sin^2x=1\)
\(\Leftrightarrow2sinx\left(cosx-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=cosx\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
a/
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin5x-\frac{1}{2}cos5x=-1\)
\(\Leftrightarrow sin\left(5x-\frac{\pi}{6}\right)=-1\)
\(\Leftrightarrow5x-\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=-\frac{\pi}{15}+\frac{k2\pi}{5}\)
b/
\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(1.\hept{\begin{cases}2-2\cos x\ge0\\\sqrt{2-2\cos x}-2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}\cos x\le1\left(đ\right)\\\cos x\ne-1\end{cases}}\Leftrightarrow x\ne\pi+k2\pi\left(k\in Z\right)\)
\(2.\hept{\begin{cases}\sin3x\ne0\\1+\sin3x\ge0\\1-\sqrt{1+\sin3x}\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x\ne k\pi\\\sin3x\ge-1\left(đ\right)\\\sin3x\ne0\end{cases}}\Leftrightarrow x\ne\frac{k\pi}{3}\left(k\in Z\right)\)
\(3.\hept{\begin{cases}\sin2x\ne0\\\sin x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\ne k\pi\\x\ne k\pi\end{cases}}\Leftrightarrow x\ne\frac{k\pi}{2}\left(k\in Z\right)\)
c/
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=cos3x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos3x\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=3x+k2\pi\\x+\frac{\pi}{3}=-3x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{12}+\frac{k\pi}{2}\end{matrix}\right.\)
d/
\(\Leftrightarrow\frac{1}{2}sin3x-\frac{\sqrt{3}}{2}cos3x=sin2x\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{3}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{3}=2x+k2\pi\\3x-\frac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{15}+\frac{k2\pi}{5}\end{matrix}\right.\)
a/
\(\Leftrightarrow\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx=sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{6}\right)\)
\(\Rightarrow x+\frac{\pi}{3}=\pi-x-\frac{\pi}{6}+k2\pi\)
\(\Rightarrow x=\frac{\pi}{4}+k\pi\)
b/
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx=sin\frac{\pi}{12}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{6}\right)=sin\frac{\pi}{12}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=\frac{\pi}{12}+k2\pi\\x+\frac{\pi}{6}=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+k2\pi\\x=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
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.
1.
\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)+sinx.cosx-1=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)-\left(1-sinx.cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx-1\right)\left(1-sinx.cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=1\\sinx.cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\\\frac{1}{2}sin2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\\sin2x=2\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
\(\Leftrightarrow\frac{1}{2}cosx+\frac{\sqrt{3}}{2}sinx=cos2x\)
\(\Leftrightarrow cos2x=cos\left(x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=x-\frac{\pi}{3}+k2\pi\\2x=\frac{\pi}{3}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
3.
\(\Leftrightarrow\sqrt{3}cosx-3sinx=2sin5x-2sinx\)
\(\Leftrightarrow\sqrt{3}cosx-sinx=2sin5x\)
\(\Leftrightarrow-\left(\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx\right)=sin5x\)
\(\Leftrightarrow sin5x=-sin\left(x-\frac{\pi}{3}\right)=sin\left(\frac{\pi}{3}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\frac{\pi}{3}-x+k2\pi\\5x=\frac{2\pi}{3}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)