Toán Lớp 11: Xin loi vi lam phien, cac cau giup minh giai cau nay duoc khong? Sinx-sinx.cosx+cosx=0
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Toán Lớp 11: Xin loi vi lam phien, cac cau giup minh giai cau nay duoc khong? Sinx-sinx.cosx+cosx=0
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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.$
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}$