Bài tập nâng cao · Bài 75
7. HỆ PHƯƠNG TRÌNH – BẤT PHƯƠNG TRÌNH LƯỢNG GIÁC
Bài tập nâng cao Phương trình lượng giác cơ bản · Bài 75
Giải các hệ phương trình sau:
1. $\left\{ \begin{array}{l} 2\cos x-1=0 \\ \sin 2x=\frac{\sqrt{3}}{2} \end{array} \right.$
2. $\left\{ \begin{array}{l} \sin x+\sin y=2 \\ \sin^{2}x+\sin^{2}y=2 \end{array} \right.$
3. $\left\{ \begin{array}{l} \sin x.\sin y=-\frac{1}{2} \\ \cos x.\cos y=\frac{1}{2} \end{array} \right.$
4. $\left\{ \begin{array}{l} \tan x+\tan y+\tan x.\tan y=1 \\ 3\sin 2y-2=\cos 4x \end{array} \right.$
1. $\left\{ \begin{array}{l} 2\cos x-1=0 \\ \sin 2x=\frac{\sqrt{3}}{2} \end{array} \right.$
2. $\left\{ \begin{array}{l} \sin x+\sin y=2 \\ \sin^{2}x+\sin^{2}y=2 \end{array} \right.$
3. $\left\{ \begin{array}{l} \sin x.\sin y=-\frac{1}{2} \\ \cos x.\cos y=\frac{1}{2} \end{array} \right.$
4. $\left\{ \begin{array}{l} \tan x+\tan y+\tan x.\tan y=1 \\ 3\sin 2y-2=\cos 4x \end{array} \right.$
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Lời giải
1. $\left\{ \begin{array}{l} 2\cos x-1=0 \\ \sin 2x=\frac{\sqrt{3}}{2} \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} 2\cos x=1 \\ \sin 2x=\frac{\sqrt{3}}{2} \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} x=\pm\frac{\pi}{3}+k2\pi \\ \left[ \begin{array}{l} x=\frac{\pi}{6}+l\pi \\ x=\frac{\pi}{3}+l\pi \end{array} \right. \end{array} \right.$ $\Leftrightarrow x=\frac{\pi}{3}+k2\pi\left( k\in\mathbb{Z} \right)$
2. $\left\{ \begin{array}{l} \sin x+\sin y=2 \\ \sin^{2}x+\sin^{2}y=2 \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} \sin x=1 \\ \sin y=1 \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} x=\frac{\pi}{2}+k2\pi \\ y=\frac{\pi}{2}+k2\pi \end{array} \right.\left( k\in\mathbb{Z} \right)$
3. $\left\{ \begin{array}{l} \sin x.\sin y=-\frac{1}{2} \\ \cos x.\cos y=\frac{1}{2} \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} \frac{1}{2}\left[ \cos\left( x-y \right)-\cos\left( x+y \right) \right]=-\frac{1}{2} \\ \cos x.\cos y+\sin x.\sin y=0 \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} \cos\left( x-y \right)-\cos\left( x+y \right)=-1 \\ \cos\left( x-y \right)=0 \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} \cos\left( x+y \right)=1 \\ \cos\left( x-y \right)=0 \end{array} \right.$ $\left\{ \begin{array}{l} x+y=k\pi 2 \\ x-y=\frac{\pi}{2}+l\pi \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} x=\frac{\pi}{4}+k\pi+\frac{l\pi}{2} \\ y=k\pi-\frac{\pi}{4}-\frac{l\pi}{2} \end{array} \right.$
4. $\left\{ \begin{array}{l} \tan x+\tan y+\tan x.\tan y=1 \\ 3\sin 2y-2=\cos 4x \end{array} \right.$
Ta có: $\tan x+\tan y+\tan x.\tan y=1$ $\Leftrightarrow \tan x+\tan y=1-\tan x\tan y$
$\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} \frac{\tan x+\tan y}{1-\tan x\tan y}=1 \\ 1-\tan x\tan y\neq 0 \end{array} \right. \\ \left\{ \begin{array}{l} 1-\tan x\tan y=0 \\ \tan x+\tan y=0 \end{array} \right. \end{array} \right.$ $\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} \tan\left( x+y \right)=1 \\ 1-\tan x\tan y\neq 0 \end{array} \right. \\ \left\{ \begin{array}{l} \tan x\tan y=1 \\ \tan x+\tan y=0 \end{array} \right.\text{(VN)} \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} x+y=\frac{\pi}{4}+k\pi \\ 1-\tan x\tan y\neq 0 \end{array} \right.$ $\Leftrightarrow x+y=\frac{\pi}{4}+k\pi$
$\Rightarrow x=\frac{\pi}{4}-y+k\pi$
$\Rightarrow 3\sin 2y-2=\cos 4x$ $\Leftrightarrow 3\sin 2y-2=\cos\left( \pi-4y \right)=-\cos 4y$
$\Leftrightarrow 3\sin 2y-2=2\sin^{2}2y-1$ $\Leftrightarrow 2\sin^{2}2y-3\sin 2y+1=0$ $\Leftrightarrow \left[ \begin{array}{l} \sin 2y=1 \\ \sin 2y=\frac{1}{2} \end{array} \right.$ $\Leftrightarrow \left[ \begin{array}{l} y=\frac{\pi}{4}+k\pi\Rightarrow x=k\pi \\ y=\frac{\pi}{12}+k\pi\Rightarrow x=\frac{\pi}{6}+k\pi \\ y=\frac{5\pi}{12}+k\pi\Rightarrow x=-\frac{\pi}{6}+k\pi \end{array} \right.$
2. $\left\{ \begin{array}{l} \sin x+\sin y=2 \\ \sin^{2}x+\sin^{2}y=2 \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} \sin x=1 \\ \sin y=1 \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} x=\frac{\pi}{2}+k2\pi \\ y=\frac{\pi}{2}+k2\pi \end{array} \right.\left( k\in\mathbb{Z} \right)$
3. $\left\{ \begin{array}{l} \sin x.\sin y=-\frac{1}{2} \\ \cos x.\cos y=\frac{1}{2} \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} \frac{1}{2}\left[ \cos\left( x-y \right)-\cos\left( x+y \right) \right]=-\frac{1}{2} \\ \cos x.\cos y+\sin x.\sin y=0 \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} \cos\left( x-y \right)-\cos\left( x+y \right)=-1 \\ \cos\left( x-y \right)=0 \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} \cos\left( x+y \right)=1 \\ \cos\left( x-y \right)=0 \end{array} \right.$ $\left\{ \begin{array}{l} x+y=k\pi 2 \\ x-y=\frac{\pi}{2}+l\pi \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} x=\frac{\pi}{4}+k\pi+\frac{l\pi}{2} \\ y=k\pi-\frac{\pi}{4}-\frac{l\pi}{2} \end{array} \right.$
4. $\left\{ \begin{array}{l} \tan x+\tan y+\tan x.\tan y=1 \\ 3\sin 2y-2=\cos 4x \end{array} \right.$
Ta có: $\tan x+\tan y+\tan x.\tan y=1$ $\Leftrightarrow \tan x+\tan y=1-\tan x\tan y$
$\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} \frac{\tan x+\tan y}{1-\tan x\tan y}=1 \\ 1-\tan x\tan y\neq 0 \end{array} \right. \\ \left\{ \begin{array}{l} 1-\tan x\tan y=0 \\ \tan x+\tan y=0 \end{array} \right. \end{array} \right.$ $\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} \tan\left( x+y \right)=1 \\ 1-\tan x\tan y\neq 0 \end{array} \right. \\ \left\{ \begin{array}{l} \tan x\tan y=1 \\ \tan x+\tan y=0 \end{array} \right.\text{(VN)} \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} x+y=\frac{\pi}{4}+k\pi \\ 1-\tan x\tan y\neq 0 \end{array} \right.$ $\Leftrightarrow x+y=\frac{\pi}{4}+k\pi$
$\Rightarrow x=\frac{\pi}{4}-y+k\pi$
$\Rightarrow 3\sin 2y-2=\cos 4x$ $\Leftrightarrow 3\sin 2y-2=\cos\left( \pi-4y \right)=-\cos 4y$
$\Leftrightarrow 3\sin 2y-2=2\sin^{2}2y-1$ $\Leftrightarrow 2\sin^{2}2y-3\sin 2y+1=0$ $\Leftrightarrow \left[ \begin{array}{l} \sin 2y=1 \\ \sin 2y=\frac{1}{2} \end{array} \right.$ $\Leftrightarrow \left[ \begin{array}{l} y=\frac{\pi}{4}+k\pi\Rightarrow x=k\pi \\ y=\frac{\pi}{12}+k\pi\Rightarrow x=\frac{\pi}{6}+k\pi \\ y=\frac{5\pi}{12}+k\pi\Rightarrow x=-\frac{\pi}{6}+k\pi \end{array} \right.$