Phương trình lượng giác.(I)

Question

$a)\,2\tan x+\cot x=2\sin2x+\dfrac{1}{\sin2x}\\ b)\,\sqrt{3}\sin\left(3x-\dfrac{\pi}{5}\right)+2\sin\left(8x-\dfrac{\pi}{3}\right)=2\sin\left(2x+\dfrac{11\pi}{15}\right)+3\cos\left(3x-\dfrac{\pi}{5}\right)\\c)\,\sin^2x+\sin x\cos3x+\cos^23x=\dfrac{3}{4}\\d)\,\left(\tan x\cot2x-1\right)\sin\left(4x+\dfrac{\pi}{2}\right)=-\dfrac{1}{2}\left(\sin^4x+\cos^4x\right)$

score 4 · Answer 1

d)
Điều kiện: $\left\{ \begin{array}{l} \cos x\ne0\\\sin2x\ne0 \end{array} \right.\Leftrightarrow x\ne\frac{k\pi}{2}$
Phương trình tương đương với:
$\left(\frac{\sin x\cos2x}{\cos x\sin2x}-1\right)\cos4x=-\frac{1}{2}\left((\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x\right)$
$\Leftrightarrow \frac{-\sin x}{\cos x\sin2x}\cos4x=-\frac{1}{2}\left(1-\frac{1}{2}\sin^22x\right)$
$\Leftrightarrow \frac{\cos4x}{\cos^2x}=\frac{1+\cos^22x}{2}$
$\Leftrightarrow \frac{2(2\cos^2x-1)^2-1}{\cos^2x}=\frac{1+(2\cos^2x-1)^2}{2}$
$\Leftrightarrow 2\cos^6x-10\cos^4x+9\cos^2x-1=0$
$\Leftrightarrow \left[ \begin{array}{l} \cos^2x=1&\textrm{(loại)}\\\cos^2x=\frac{4+\sqrt{14}}{2}&\textrm{(loại)}\\\cos^2x=\frac{4-\sqrt{14}}{2} \end{array} \right.$
$\Leftrightarrow \cos 2x=14-4\sqrt{14}$
$\Leftrightarrow x=\pm\frac{1}{2}\arccos(14-4\sqrt{14})+k\pi,k\in\mathbb{Z}$.

score 4 · Answer 2

c) Phương trình tương đương với:
$\sin^2x+\sin x\cos3x+\cos^23x=\frac{3}{4}(\sin^2x+\cos^2x)$
$\Leftrightarrow \left(\frac{1}{2}\sin x+\cos3x\right)^2=\left(\frac{\sqrt3}{2}\cos x\right)^2$
$\Leftrightarrow \left[ \begin{array}{l} \frac{1}{2}\sin x+\cos3x=\frac{\sqrt3}{2}\cos x\\\frac{1}{2}\sin x+\cos3x=-\frac{\sqrt3}{2}\cos x\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} \cos 3x=\cos\left(x+\frac{\pi}{6}\right)\\\cos(3x+\pi)=\cos\left(x-\frac{\pi}{6}\right) \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} 3x=x+\frac{\pi}{6}+k2\pi\\3x=-x-\frac{\pi}{6}+k2\pi\\3x+\pi=x-\frac{\pi}{6}+k2\pi\\3x+\pi=-x+\frac{\pi}{6}+k2\pi \end{array} \right.,k\in\mathbb{Z}\Leftrightarrow \left[ \begin{array}{l} x=\frac{\pi}{12}+k\pi\\ x=\frac{-\pi}{24}+\frac{k\pi}{2}\\x=\frac{-7\pi}{12}+k\pi\\x=\frac{-5\pi}{24}+\frac{k\pi}{2}\end{array} \right.,k\in\mathbb{Z}$

score 4 · Answer 3

a) Điều kiện: $\displaystyle \sin2x\ne0\Leftrightarrow x\ne\frac{k\pi}{2},k\in\mathbb{Z}$.
Phương trình tương đương với:
$\displaystyle 2(\tan x-\sin 2x)+\left(\frac{1}{\tan x}-\frac{1}{\sin 2x}\right)=0$
$\displaystyle \Leftrightarrow (\tan x-\sin2x)\left(2-\frac{1}{\tan x\sin 2x}\right)=0$
$\displaystyle \Leftrightarrow (\tan x-\sin2x)\left(2-\frac{1}{2\sin^2x}\right)=0$
$\displaystyle \Leftrightarrow \left[ \begin{array}{l} \tan x=\sin2x\\ 4\sin^2x=1 \end{array} \right.$
$\displaystyle \Leftrightarrow \left[ \begin{array}{l} \tan x=\frac{2\tan x}{1+\tan^2x}\\2(1-\cos 2x)=1 \end{array} \right.$
$\displaystyle \Leftrightarrow \left[ \begin{array}{l} \tan x=0\\\tan x=1\\\tan x=-1\\\cos2x=\frac{1}{2} \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=k\pi\\x=\pm\frac{\pi}{4}+k\pi\\x=\pm\frac{\pi}{6}+k\pi \end{array} \right.,k\in\mathbb{Z}$.
Kết hợp với điều kiện ta có: $\displaystyle x\in\{\pm\frac{\pi}{4}+k\pi,\pm\frac{\pi}{6}+k\pi,k\in\mathbb{Z}\}$