Bài tập nâng cao · Bài 41
4. PHƯƠNG TRÌNH ĐỐI XỨNG Nhận dạng: $a\left( \sin x\pm \cos x \right)+b\sin x\cos x=c$. Cách làm: Đặt $t=\sin x+\cos x\Rightarrow \sin x\cos x=\frac{t^{2} -1}{2}$. Điều kiện $\left| t \right|\leq \sqrt{2}$. $t=\sin x-\cos x\Rightarrow \sin x\cos x=\frac{1-t^{2}}{2}$. Điều kiện $\left| t \right|\leq \sqrt{2}$.
Bài tập nâng cao Phương trình lượng giác cơ bản · Bài 41
Giải các phương trình sau:
a. $\sin 2x-12\left( \sin x-\cos x \right)+12=0$.
b. $\sin x\cos x+2\left( \sin x+\cos x \right)=2$.
c. $\sin x+\cos x=\frac{2\sqrt{3}}{3}\sqrt{1+\sin x\cos x}$.
d. $\sin x-\cos x+7\sin 2x=1$.
a. $\sin 2x-12\left( \sin x-\cos x \right)+12=0$.
b. $\sin x\cos x+2\left( \sin x+\cos x \right)=2$.
c. $\sin x+\cos x=\frac{2\sqrt{3}}{3}\sqrt{1+\sin x\cos x}$.
d. $\sin x-\cos x+7\sin 2x=1$.
Xem lời giải
Lời giải
a. $\sin 2x-12\left( \sin x-\cos x \right)+12=0$.
Đặt $t=\sin x-\cos x \left( -\sqrt{2}\leq t\leq \sqrt{2} \right)\Rightarrow t^{2}=1-\sin 2x\Rightarrow \sin 2x=1-t^{2}$.
PT $\Leftrightarrow 1-t^{2}-12t+12=0\Leftrightarrow -t^{2}-12t+13=0\Leftrightarrow \left[ \begin{array}{l} t=1 \left( tm \right) \\ t=-13 \left( l \right) \end{array} \right.$.
$\Rightarrow \sin x-\cos x=1\Leftrightarrow \sin\left( x-\frac{\pi}{4} \right)=\frac{\sqrt{2}}{2}\Leftrightarrow \left[ \begin{array}{l} x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi \\ x-\frac{\pi}{4}=\pi -\frac{\pi}{4}+k2\pi \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=\frac{\pi}{2}+k2\pi \\ x=\pi +k2\pi \end{array} \right.$.
b. $\sin x\cos x+2\left( \sin x+\cos x \right)=2$.
Đặt $t=\sin x+\cos x \left( -\sqrt{2}\leq t\leq \sqrt{2} \right)\Rightarrow t^{2}=1+2\sin x\cos x\Rightarrow \sin x\cos x=\frac{t^{2}-1}{2}$.
PT $\Leftrightarrow \frac{t^{2}-1}{2}+2t-2=0\Leftrightarrow t^{2}+4t-5=0\Leftrightarrow \left[ \begin{array}{l} t=1 \left( tm \right) \\ t=-5 \left( l \right) \end{array} \right.$.
$\Rightarrow \sin x+\cos x=1\Leftrightarrow \sin\left( x+\frac{\pi}{4} \right)=\frac{\sqrt{2}}{2}\Leftrightarrow \left[ \begin{array}{l} x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi \\ x+\frac{\pi}{4}=\pi -\frac{\pi}{4}+k2\pi \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=k2\pi \\ x=\frac{\pi}{2}+k2\pi \end{array} \right.$.
c. $\sin x+\cos x=\frac{2\sqrt{3}}{3}\sqrt{1+\sin x\cos x}$.
Đặt $t=\sin x+\cos x \left( -\sqrt{2}\leq t\leq \sqrt{2} \right)\Rightarrow t^{2}=1+2\sin x\cos x\Rightarrow \sin x\cos x=\frac{t^{2}-1}{2}$.
PT $\Leftrightarrow t=\frac{2\sqrt{3}}{3}\sqrt{1+\frac{t^{2}-1}{2}}\Leftrightarrow \left\{ \begin{array}{l} t\geq 0 \\ t^{2}=\frac{4}{3}.\frac{t^{2}+1}{2}\Leftrightarrow t^{2}=2 \end{array} \right.\Leftrightarrow t=\sqrt{2}$.
$\Rightarrow \sin x+\cos x=\sqrt{2}\Leftrightarrow \sin\left( x+\frac{\pi}{4} \right)=1\Leftrightarrow x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi \Leftrightarrow x=\frac{\pi}{4}+k2\pi$.
d. $\sin x-\cos x+7\sin 2x=1$.
Đặt $t=\sin x-\cos x \left( -\sqrt{2}\leq t\leq \sqrt{2} \right)\Rightarrow t^{2}=1-\sin 2x\Rightarrow \sin 2x=1-t^{2}$.
PT $\Leftrightarrow t+7\left( 1-t^{2} \right)=1\Leftrightarrow -7t^{2}+t+6=0\Leftrightarrow \left[ \begin{array}{l} t=1 \left( tm \right) \\ t=-\frac{6}{7} \left( tm \right) \end{array} \right.$.
$\left[ \begin{array}{l} t=1 \\ t=-\frac{6}{7} \end{array} \right.\Rightarrow \left[ \begin{array}{l} \sin x-\cos x=1 \\ \sin x-\cos x=-\frac{6}{7} \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} \sin\left( x-\frac{\pi}{4} \right)=\frac{\sqrt{2}}{2} \\ \sin\left( x-\frac{\pi}{4} \right)=\sin\alpha \left( \sin\alpha =-\frac{3\sqrt{2}}{7} \right) \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi \\ x-\frac{\pi}{4}=\pi -\frac{\pi}{4}+k2\pi \\ x-\frac{\pi}{4}=\alpha +k2\pi \\ x-\frac{\pi}{4}=\pi -\alpha +k2\pi \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=\frac{\pi}{2}+k2\pi \\ x=\pi +k2\pi \\ x=\alpha +\frac{\pi}{4}+k2\pi \\ x=\frac{5\pi}{4}-\alpha +k2\pi \end{array} \right.$.
Đặt $t=\sin x-\cos x \left( -\sqrt{2}\leq t\leq \sqrt{2} \right)\Rightarrow t^{2}=1-\sin 2x\Rightarrow \sin 2x=1-t^{2}$.
PT $\Leftrightarrow 1-t^{2}-12t+12=0\Leftrightarrow -t^{2}-12t+13=0\Leftrightarrow \left[ \begin{array}{l} t=1 \left( tm \right) \\ t=-13 \left( l \right) \end{array} \right.$.
$\Rightarrow \sin x-\cos x=1\Leftrightarrow \sin\left( x-\frac{\pi}{4} \right)=\frac{\sqrt{2}}{2}\Leftrightarrow \left[ \begin{array}{l} x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi \\ x-\frac{\pi}{4}=\pi -\frac{\pi}{4}+k2\pi \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=\frac{\pi}{2}+k2\pi \\ x=\pi +k2\pi \end{array} \right.$.
b. $\sin x\cos x+2\left( \sin x+\cos x \right)=2$.
Đặt $t=\sin x+\cos x \left( -\sqrt{2}\leq t\leq \sqrt{2} \right)\Rightarrow t^{2}=1+2\sin x\cos x\Rightarrow \sin x\cos x=\frac{t^{2}-1}{2}$.
PT $\Leftrightarrow \frac{t^{2}-1}{2}+2t-2=0\Leftrightarrow t^{2}+4t-5=0\Leftrightarrow \left[ \begin{array}{l} t=1 \left( tm \right) \\ t=-5 \left( l \right) \end{array} \right.$.
$\Rightarrow \sin x+\cos x=1\Leftrightarrow \sin\left( x+\frac{\pi}{4} \right)=\frac{\sqrt{2}}{2}\Leftrightarrow \left[ \begin{array}{l} x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi \\ x+\frac{\pi}{4}=\pi -\frac{\pi}{4}+k2\pi \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=k2\pi \\ x=\frac{\pi}{2}+k2\pi \end{array} \right.$.
c. $\sin x+\cos x=\frac{2\sqrt{3}}{3}\sqrt{1+\sin x\cos x}$.
Đặt $t=\sin x+\cos x \left( -\sqrt{2}\leq t\leq \sqrt{2} \right)\Rightarrow t^{2}=1+2\sin x\cos x\Rightarrow \sin x\cos x=\frac{t^{2}-1}{2}$.
PT $\Leftrightarrow t=\frac{2\sqrt{3}}{3}\sqrt{1+\frac{t^{2}-1}{2}}\Leftrightarrow \left\{ \begin{array}{l} t\geq 0 \\ t^{2}=\frac{4}{3}.\frac{t^{2}+1}{2}\Leftrightarrow t^{2}=2 \end{array} \right.\Leftrightarrow t=\sqrt{2}$.
$\Rightarrow \sin x+\cos x=\sqrt{2}\Leftrightarrow \sin\left( x+\frac{\pi}{4} \right)=1\Leftrightarrow x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi \Leftrightarrow x=\frac{\pi}{4}+k2\pi$.
d. $\sin x-\cos x+7\sin 2x=1$.
Đặt $t=\sin x-\cos x \left( -\sqrt{2}\leq t\leq \sqrt{2} \right)\Rightarrow t^{2}=1-\sin 2x\Rightarrow \sin 2x=1-t^{2}$.
PT $\Leftrightarrow t+7\left( 1-t^{2} \right)=1\Leftrightarrow -7t^{2}+t+6=0\Leftrightarrow \left[ \begin{array}{l} t=1 \left( tm \right) \\ t=-\frac{6}{7} \left( tm \right) \end{array} \right.$.
$\left[ \begin{array}{l} t=1 \\ t=-\frac{6}{7} \end{array} \right.\Rightarrow \left[ \begin{array}{l} \sin x-\cos x=1 \\ \sin x-\cos x=-\frac{6}{7} \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} \sin\left( x-\frac{\pi}{4} \right)=\frac{\sqrt{2}}{2} \\ \sin\left( x-\frac{\pi}{4} \right)=\sin\alpha \left( \sin\alpha =-\frac{3\sqrt{2}}{7} \right) \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi \\ x-\frac{\pi}{4}=\pi -\frac{\pi}{4}+k2\pi \\ x-\frac{\pi}{4}=\alpha +k2\pi \\ x-\frac{\pi}{4}=\pi -\alpha +k2\pi \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=\frac{\pi}{2}+k2\pi \\ x=\pi +k2\pi \\ x=\alpha +\frac{\pi}{4}+k2\pi \\ x=\frac{5\pi}{4}-\alpha +k2\pi \end{array} \right.$.