Mn giúp em pt này: Sin^2(x) – Cos^2(2x)=0 30/07/2021 Bởi Serenity Mn giúp em pt này: Sin^2(x) – Cos^2(2x)=0
$\sin^2x – \cos^22x = 0$ $\Leftrightarrow\dfrac{1 – \cos2x}{2} – \cos^22x = 0$ $\Leftrightarrow 2\cos^22x + \cos2x – 1 = 0$ $\Leftrightarrow\left[\begin{array}{l}\cos2x = -1\\\cos2x =\dfrac{1}{2}\end{array}\right.$ $\Leftrightarrow \left[\begin{array}{l}2x = \pi + k2\pi\\2x =\pm \dfrac{\pi}{3} + k2\pi\end{array}\right.$ $\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k\pi\\x =\pm \dfrac{\pi}{6} + k\pi\end{array}\right.\qquad (k\in \Bbb Z)$ Bình luận
$\sin^2x – \cos^22x = 0$
$\Leftrightarrow\dfrac{1 – \cos2x}{2} – \cos^22x = 0$
$\Leftrightarrow 2\cos^22x + \cos2x – 1 = 0$
$\Leftrightarrow\left[\begin{array}{l}\cos2x = -1\\\cos2x =\dfrac{1}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}2x = \pi + k2\pi\\2x =\pm \dfrac{\pi}{3} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k\pi\\x =\pm \dfrac{\pi}{6} + k\pi\end{array}\right.\qquad (k\in \Bbb Z)$
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