22 câu Dạng 4. Phương trình lượng giác đối xứng

Đặt $t=\sin x-\cos x\left(-\sqrt{2}\le t\le \sqrt{2}\right)\Rightarrow \sin x\cos x=\frac{1-t^2}{2}.$

Khi đó phương trình (1) trở thành $t+7\left(1-t^2\right)=1\iff 7t^2-t-6=0\iff \left[\begin{array}{l}t=1\\ t=-\frac{6}{7}\end{array}\right..$

- Nếu $t=1$  thì $\sin x-\cos x=1\iff \sin \left(x-\frac{\pi }{4}\right)=\sin \frac{\pi }{4}\iff \left[\begin{array}{l}x=\frac{\pi }{2}+k2\pi \\ x=\pi +k2\pi \end{array}\right.\left(k\in \mathbb{Z}\right).$

- Nếu $t=-\frac{6}{7}$  thì $\sin x-\cos x=\frac{-6}{7}$

$\iff \sin \left(x-\frac{\pi }{4}\right)=\frac{-3\sqrt{2}}{7}\iff \left[\begin{array}{l}x=\frac{\pi }{4}+\arcsin \frac{-3\sqrt{2}}{7}+k2\pi \\ x=\frac{5\pi }{4}-\arcsin \frac{-3\sqrt{2}}{7}+k2\pi \end{array}\right.\left(k\in \mathbb{Z}\right).$

Vậy phương trình đã cho có 4 họ nghiệm $x=\frac{\pi }{2}+k2\pi ;x=\pi +k2\pi$

$x=\frac{\pi }{4}+\arcsin \frac{-3\sqrt{2}}{7}+k2\pi ;x=\frac{5\pi }{4}-\arcsin \frac{-3\sqrt{2}}{7}+k2\pi \left(k\in \mathbb{Z}\right).$

$\left(2\right)\iff 1+\left(\sin x+\cos x\right)\left({\sin }^{2}x-\sin x\cos x+{\cos }^{2}x\right)=3\sin x\cos x$

$\iff 1+\left(\sin x+\cos x\right)\left(1-\sin x\cos x\right)=3\sin x\cos x.\left(*\right)$

Đặt $t=\sin x+\cos x\left(-\sqrt{2}\le t\le \sqrt{2}\right)\Rightarrow \sin x\cos x=\frac{t^2-1}{2}.$

Khi đó phương trình (*) trở thành $1+t\left(1-\frac{t^2-1}{2}\right)=3.\frac{t^2-1}{2}$

$\iff t^3+3t^2-3t-5=0\iff \left(t+1\right)\left(t^2+2t-5\right)=0\iff \left[\begin{array}{l}t=-1\\ t=-1-\sqrt{6}<-\sqrt{2}\\ t=-1+\sqrt{6}>\sqrt{2}\end{array}\right.\Rightarrow t=-1.$

Suy ra $\sin x+\cos x=-1\iff \sqrt{2}cos\left(x-\frac{\pi }{4}\right)=-1$

                            $\iff cos\left(x-\frac{\pi }{4}\right)=cos\frac{3\pi }{4}\iff \left[\begin{array}{l}x=\pi +k2\pi \\ x=\frac{-\pi }{2}+k2\pi \end{array}\right.\left(k\in \mathbb{Z}\right).$             

Vậy phương trình đã cho có 2 họ nghiệm $x=\pi +k2\pi ;x=\frac{-\pi }{2}+k2\pi \left(k\in \mathbb{Z}\right).$

Phương trình  $-\sqrt{2}\left(\sin x+\cos x\right)+2\sin x\cos x+1=0\left(1\right)$ có nghĩa $\forall x\in ℝ\iff D=ℝ.$

Đặt $t=\sin x+\cos x,\left(\left|t\right|\le \sqrt{2}\right).$

Đáp án D

Phương trình $\left(1+\sin x\right)\left(1+\cos x\right)=2$  có nghĩa $\forall x\in ℝ\iff D=ℝ.$

Ta có $\left(1+\sin x\right)\left(1+\cos x\right)=2\iff \cos x+\sin x+\sin x\cos x=1.$

Đặt $t=\sin x+\cos x,\left(\left|t\right|\le \sqrt{2}\right).$

Ta có $\sin x\cos x=\frac{t^2-1}{2}\left(1\right)\Rightarrow t^2+2t-3=0\iff \left[\begin{array}{l}t=1\\ t=-3\end{array}\right..$

Do  $\left|t\right|\le \sqrt{2}$ nên $t=1.$

Với $t=\mathrm{1,}$  ta có $t=\sin x+\cos x=\sqrt{2}\cos \left(x-\frac{\pi }{4}\right)=1\iff \cos \left(x-\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}.$

Đáp án A

Phương trình $\sin x-\cos x+2\sin 2x+1=0$  có nghĩa $\forall x\in ℝ\iff D=ℝ.$

Ta có $\sin x-\cos x+2\sin 2x+1=0\iff \sin x-\cos x+4\sin x\cos x+1=0.\left(1\right)$

Đặt $t=\sin x-\cos x,\left(\left|t\right|\le \sqrt{2}\right).$

Ta có $\sin x\cos x=\frac{1-t^2}{2}\Rightarrow \left(1\right)\iff t+2\left(1-t^2\right)+1=0\iff 2t^2-t-3=0\iff \left[\begin{array}{l}t=-1\\ t=\frac{3}{2}\end{array}\right..$

Do  $\left|t\right|\le \sqrt{2}$ nên $t=-1.$

Với $t=-\mathrm{1,}$  ta có $t=\sin x-\cos x=\sqrt{2}\sin \left(x-\frac{\pi }{4}\right)=-1\iff \sin \left(x-\frac{\pi }{4}\right)=-\frac{1}{\sqrt{2}}=\sin \frac{-\pi }{4}$

$\iff \left[\begin{array}{l}x-\frac{\pi }{4}=-\frac{\pi }{4}+k2\pi \iff x=k2\pi \\ x-\frac{\pi }{4}=\pi -\frac{-\pi }{4}+k2\pi \iff x=\frac{3\pi }{2}+k2\pi \end{array}\right.,k\in \mathbb{Z}.$

Đáp án C

Phương trình  $\sin 2x-2\left(\sin x-\cos x\right)-2=0$ có nghĩa $\forall x\in ℝ\iff D=ℝ.$

Ta có $\sin 2x-2\left(\sin x-\cos x\right)-2=0\iff 2\left(\sin x-\cos x\right)-2\sin x\cos x+2=0.\left(1\right)$

Đặt  $t=\sin x-\cos x,\left(\left|t\right|\le \sqrt{2}\right).$ Ta có $\sin x\cos x=\frac{1-t^2}{2}$

$\Rightarrow \left(1\right)\iff 2t-\left(1-t^2\right)+2=0\iff t^2+2t+1=0\iff t=-1.$

Với $t=-\mathrm{1,}$  ta có $t=\sin x-\cos x=\sqrt{2}\sin \left(x-\frac{\pi }{4}\right)=-1\iff \sin \left(x-\frac{\pi }{4}\right)=-\frac{1}{\sqrt{2}}=\sin \frac{-\pi }{4}$

$\iff \left[\begin{array}{l}x-\frac{\pi }{4}=-\frac{\pi }{4}+k2\pi \iff x=k2\pi \\ x-\frac{\pi }{4}=\pi -\frac{-\pi }{4}+k2\pi \iff x=\frac{3\pi }{2}+k2\pi \end{array}\right.,k\in \mathbb{Z}.$

Vậy nghiệm dương nhỏ nhất của phương trình là   $x=\frac{3\pi }{2}.$

Đáp án A

Phương trình $\sin 2x+2\left(\cos x-\sin x\right)-1=0$  có nghĩa $\forall x\in ℝ\iff D=ℝ.$

Ta có $\sin 2x+2\left(\cos x-\sin x\right)-1=0\iff 2\sin x\cos x-2\left(\sin x-\cos x\right)-1=0.\left(1\right)$

Đặt  $t=\sin x-\cos x,\left(\left|t\right|\le \sqrt{2}\right).$ Ta có $\sin x\cos x=\frac{1-t^2}{2}$

 $\Rightarrow \left(1\right)\iff 1-t^2-2t-1=0\iff t^2+2t=0\iff \left[\begin{array}{l}t=0\\ t=-2\end{array}\right..$

Do $\left|t\right|\le \sqrt{2}$  nên $t=0.$

Với  $t=\mathrm{0,}$ ta có $t=\sin x-\cos x=\sqrt{2}\sin \left(x-\frac{\pi }{4}\right)=0\iff \sin \left(x-\frac{\pi }{4}\right)=0\iff x-\frac{\pi }{4}=k\pi \iff x=\frac{\pi }{4}+k\pi ,k\in \mathbb{Z}.$

Đáp án B

Phương trình $\sqrt{2}\left(\sin x+\cos x\right)=\tan x+\cot x$  có nghĩa $\iff \left\{\begin{array}{l}\cos x\ne 0\iff x\ne \frac{\pi }{2}+k\pi \\ \sin x\ne 0\iff x\ne k\pi \end{array}\right.\iff x\ne k\frac{\pi }{2}.$

Ta có $\sqrt{2}\left(\sin x+\cos x\right)=\tan x+\cot x$

$\iff \sqrt{2}\left(\sin x+\cos x\right)=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\iff \sqrt{2}\left(\sin x+\cos x\right)=\frac{1}{\sin x\cos x}.\left(1\right)$

Đặt $t=\sin x+\cos x,\left(\left|t\right|\le \sqrt{2}\right).$  Ta có $\sin x\cos x=\frac{t^2-1}{2}$

$\Rightarrow \left(1\right)\iff \sqrt{2}t=\frac{2}{t^2-1}\iff \sqrt{2}{t}^3-\sqrt{2}t-2=0\left(t\ne 1\right)\iff t=\sqrt{2}.$