Toán Lớp 11: thầy cho bài tập mà hôm đó em nghĩ không biết làm, mng giải giúp e với ạ:
a) √ 3.cos(x+ π/2) – sin(x – π/2)= 2sin2x
b) cos2x + √3 sin2x= √3.cosx – sinx
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Toán Lớp 11: thầy cho bài tập mà hôm đó em nghĩ không biết làm, mng giải giúp e với ạ:
a) √ 3.cos(x+ π/2) – sin(x – π/2)= 2sin2x
b) cos2x + √3 sin2x= √3.cosx – sinx
Comments ( 1 )
a)\quad\left[\begin{array}{l}x = \dfrac{5\pi}{6} + k2\pi\\x = \dfrac{\pi}{12} + k\pi\end{array}\right.\quad (k\in\Bbb Z)\\
b)\quad \left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x =\dfrac{\pi}{18} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)
\end{array}\)
a)\quad \sqrt3\cos\left(x + \dfrac{\pi}{2}\right) – \sin\left(x – \dfrac{\pi}{2}\right) = 2\sin2x\\
\Leftrightarrow -\sqrt3\sin x + \cos x = 2\sin2x\\
\Leftrightarrow – \dfrac{\sqrt3}{2}\sin x + \dfrac12\cos x = \sin2x\\
\Leftrightarrow \sin x. \cos\dfrac{5\pi}{6} + \cos x.\sin\dfrac{5\pi}{6} = \sin2x\\
\Leftrightarrow \sin\left(x + \dfrac{5\pi}{6}\right) = \sin2x\\
\Leftrightarrow \left[\begin{array}{l}2x = x + \dfrac{5\pi}{6} + k2\pi\\2x = \pi – x – \dfrac{5\pi}{6} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{5\pi}{6} + k2\pi\\x = \dfrac{\pi}{12} + k\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy phương trình có các họ nghiệm là}\ x = \dfrac{5\pi}{6} + k2\pi\ \text{và}\ x = \dfrac{\pi}{12} + k\pi\ \text{với}\ k\in\Bbb Z\\
b)\quad \cos2x + \sqrt3\sin2x = \sqrt3\cos x – \sin x\\
\Leftrightarrow \dfrac12\cos2x + \dfrac{\sqrt3}{2}\sin2x = \dfrac{\sqrt3}{2}\cos x – \dfrac12\sin x\\
\Leftrightarrow \cos2x.\cos\dfrac{\pi}{3} + \sin2x.\sin\dfrac{\pi}{3} = \cos x.\cos\dfrac{\pi}{6} – \sin x.\sin\dfrac{\pi}{6}\\
\Leftrightarrow \cos\left(2x – \dfrac{\pi}{3}\right) = \cos\left(x + \dfrac{\pi}{6}\right)\\
\Leftrightarrow \left[\begin{array}{l}2x – \dfrac{\pi}{3} = x + \dfrac{\pi}{6} + k2\pi\\2x – \dfrac{\pi}{3} = -x – \dfrac{\pi}{6} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x =\dfrac{\pi}{18} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy phương trình có các họ nghiệm là}\ x = \dfrac{\pi}{2} + k2\pi\ \text{và}\ x = \dfrac{\pi}{18} + k\dfrac{2\pi}{3}\ \text{với}\ k\in\Bbb Z
\end{array}\)