Xin loi vi lam phien, cac cau giup minh giai cau nay duoc khong? Sinx-sinx.cosx+cosx=0

Đáp án:$\left[ \begin{array}{l}
x =  - \dfrac{\pi }{4} + \arcsin \left( {\sqrt 2  - 1} \right) + k2\pi \\
x = \dfrac{{3\pi }}{4} - \arcsin \left( {\sqrt 2  - 1} \right) + k2\pi 
\end{array} \right.$

Giải thích các bước giải:

$\begin{array}{l}
Dat:\sin x + \cos x = a\\
 \Leftrightarrow a = \sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right)\\
Do: - 1 \le \sin \left( {x + \dfrac{\pi }{4}} \right) \le 1\\
 \Leftrightarrow  - \sqrt 2  \le \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) \le \sqrt 2 \\
 \Leftrightarrow  - \sqrt 2  \le a \le \sqrt 2 \\
 \Leftrightarrow {\sin ^2}x + 2\sin x.\cos x + {\cos ^2}x = {a^2}\\
 \Leftrightarrow \left( {{{\sin }^2}x + {{\cos }^2}x} \right) + 2.\sin x.\cos x = {a^2}\\
 \Leftrightarrow 1 + 2.\sin x.\cos x = {a^2}\\
 \Leftrightarrow \sin x.\cos x = \dfrac{{{a^2} - 1}}{2}\\
\sin x - \sin x.\cos x + \cos x = 0\\
 \Leftrightarrow a - \dfrac{{{a^2} - 1}}{2} = 0\\
 \Leftrightarrow 2a - {a^2} + 1 = 0\\
 \Leftrightarrow {a^2} - 2a - 1 = 0\\
 \Leftrightarrow {a^2} - 2a + 1 = 2\\
 \Leftrightarrow {\left( {a - 1} \right)^2} = 2\\
 \Leftrightarrow \left[ \begin{array}{l}
a = \sqrt 2  - 1\left( {tm} \right)\\
a =  - \sqrt 2  - 1\left( {ktm} \right)
\end{array} \right.\\
 \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2  - 1\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = \arcsin \left( {\sqrt 2  - 1} \right) + k2\pi \\
x + \dfrac{\pi }{4} = \pi  - \arcsin \left( {\sqrt 2  - 1} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{4} + \arcsin \left( {\sqrt 2  - 1} \right) + k2\pi \\
x = \dfrac{{3\pi }}{4} - \arcsin \left( {\sqrt 2  - 1} \right) + k2\pi 
\end{array} \right.\\
Vay\,\left[ \begin{array}{l}
x =  - \dfrac{\pi }{4} + \arcsin \left( {\sqrt 2  - 1} \right) + k2\pi \\
x = \dfrac{{3\pi }}{4} - \arcsin \left( {\sqrt 2  - 1} \right) + k2\pi 
\end{array} \right.
\end{array}$