giải pt: (sinx-cosx+1)(2sinx-cosx)=sin2x

Đáp án:

$x=k2\pi$

$x=\dfrac{3\pi}2+k2\pi$

$x= \arccos\dfrac3{\sqrt{13}}+\arcsin\dfrac1{\sqrt{13}}+k2\pi$

$x= \arccos\dfrac3{\sqrt{13}}+\pi-\arcsin\dfrac1{\sqrt{13}}+k2\pi$

$(k\in\mathbb Z)$

Giải thích các bước giải:

$(\sin x-\cos x+1)(2\sin x-\cos x)=\sin 2x$

$\Leftrightarrow (\sin x-\cos x+1)(2\sin x-\cos x)=-\left[{(\sin x-\cos x)^2-1}\right]$

$\Leftrightarrow (\sin x-\cos x+1)(2\sin x-\cos x)$

$=-(\sin x-\cos x-1)(\sin x-\cos x+1)$

$\Leftrightarrow (\sin x-\cos x+1)(2\sin x-\cos x+\sin x-\cos x-1)=0$

$\Leftrightarrow (\sin x-\cos x+1)(3\sin x-2\cos x-1)=0$

$\Leftrightarrow\left[\begin{array}{I}\sin x-\cos x+1=0\text{ (1)}\\3\sin x-2\cos x-1=0\text{ (2)}\end{array}\right.$

$\text{(1)}\Leftrightarrow \sqrt2\sin\left({x-\dfrac{\pi}{4}}\right)=-1$

$\Leftrightarrow\sin\left({x-\dfrac{\pi}{4}}\right)=-\dfrac1{\sqrt2}$

$\Leftrightarrow \left[\begin{array}{I}x-\dfrac{\pi}4=-\dfrac{\pi}4+k2\pi\\x-\dfrac{\pi}4=\pi+\dfrac{\pi}4+k2\pi\end{array}\right.\Leftrightarrow \left[\begin{array}{I}x=k2\pi\\x=\dfrac{3\pi}2+k2\pi\end{array}\right.$ $(k\in\mathbb Z)$

$\text{(2)}\Leftrightarrow 3\sin x-2\cos x=1$

$\Leftrightarrow\dfrac{3}{\sqrt{13}}\sin x-\dfrac2{\sqrt{13}}\cos x=\dfrac1{\sqrt{13}}$ (3)

Đặt $\dfrac{3}{\sqrt{13}}=\cos \alpha$ và $\dfrac{2}{\sqrt{13}}=\sin\alpha$

$(\cos^2\alpha+\sin^2\alpha=1)$

(3) $\Leftrightarrow\sin x\cos\alpha-\cos x\sin \alpha=\dfrac1{\sqrt{13}}$

$\Leftrightarrow\sin (x-\alpha)=\dfrac1{\sqrt{13}}$

$\Leftrightarrow \left[\begin{array}{I}x-\alpha=\arcsin\dfrac1{\sqrt{13}}+k2\pi\\x-\alpha=\pi-\arcsin\dfrac1{\sqrt{13}}+k2\pi\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{I}x=\alpha+\arcsin\dfrac1{\sqrt{13}}+k2\pi\\x=\alpha+\pi-\arcsin\dfrac1{\sqrt{13}}+k2\pi\end{array}\right.$ $(k\in\mathbb Z)$

Trong đó $\dfrac{3}{\sqrt{13}}=\cos \alpha$ và $\dfrac{1}{\sqrt{13}}=\sin \alpha$

Do ta đặt $\cos\alpha,\sin\alpha>0$ nên $\alpha$ thuộc góc phần tư thứ nhất nên $0<\alpha<90^o$

$\Rightarrow \alpha= \arccos\dfrac3{\sqrt{13}}$

Vậy phương trình có nghiệm là:

$x=k2\pi$

$x=\dfrac{3\pi}2+k2\pi$

$x= \arccos\dfrac3{\sqrt{13}}+\arcsin\dfrac1{\sqrt{13}}+k2\pi$

$x=\arccos\dfrac3{\sqrt{13}}+\pi-\arcsin\dfrac1{\sqrt{13}}+k2\pi$

$(k\in\mathbb Z)$.