Giải phương trình: a)sin6x+sin5x+sinx=0 b)sin6x+cos8x-cos4x=0 c)(1+cosx)(2sinx-cosx)=sin²x d)tan²x=1+cosx/1-sinx e)4cos²x-2cos2x-2cos²2x=0

0 bình luận về “Giải phương trình: a)sin6x+sin5x+sinx=0 b)sin6x+cos8x-cos4x=0 c)(1+cosx)(2sinx-cosx)=sin²x d)tan²x=1+cosx/1-sinx e)4cos²x-2cos2x-2cos²2x=0”

1. a) $\sin6x + \sin5x + \sin x = 0$

   $\Leftrightarrow \sin6x + 2\sin3x\cos2x = 0$

   $\Leftrightarrow 2\sin3x\cos3x + 2\sin3x\cos2x = 0$

   $\Leftrightarrow \sin3x(\cos3x + \cos2x) = 0$

   $\Leftrightarrow \left[\begin{array}{l}\sin3x = 0\\\cos3x = \cos(\pi – 2x)\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}3x = k\pi\\3x = \pi – 2x + k2\pi\\3x = 2x – \pi + k2\pi\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}x = k\dfrac{\pi}{3}\\x = \dfrac{\pi}{5} + k\dfrac{2\pi}{5}\\x =- \pi + k2\pi\end{array}\right. \quad (k \in \Bbb Z)$

   b) $\sin6x + \cos8x – \cos4x = 0$

   $\Leftrightarrow \sin6x – 2\sin6x\sin2x = 0$

   $\Leftrightarrow \sin6x(1 – 2\sin2x) = 0$

   $\Leftrightarrow \left[\begin{array}{l}\sin6x = 0\\\sin2x = \dfrac{1}{2}\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}6x = k\pi\\2x = \dfrac{\pi}{6}+ k2\pi\\2x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}x = k\dfrac{\pi}{6}\\x = \dfrac{\pi}{12}+ k\pi\\x = \dfrac{5\pi}{12} + k\pi\end{array}\right. \quad (k \in \Bbb Z)$

   c) $(1 + \cos x)(2\sin x – \cos x) = \sin^2x$

   $\Leftrightarrow (1 + \cos x)(2\sin x – \cos x) = 1 – \cos^2x$

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

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

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

   $\Leftrightarrow \left[\begin{array}{l}\cos x = -1\\\sin x = \dfrac{1}{2}\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}x = \pi + k2\pi\\x = \dfrac{\pi}{6}+ k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right. (k \in \Bbb Z)$

   d) $tan^2x = \dfrac{1 + \cos x}{1 – \sin x}\qquad (*)$

   $ĐK: \, \begin{cases}\cos x \ne 0\\\sin x \ne 1\end{cases} \Leftrightarrow x \ne \dfrac{\pi}{2} + k\pi$

   $(*) \Leftrightarrow \dfrac{1 – \cos^2x}{1 – \sin^2x} – \dfrac{1 +\cos x}{1 – \sin x} = 0$

   $\Leftrightarrow \dfrac{(1 – \cos x)(1 + \cos x)}{1 + \sin x} – (1 + \cos x) = 0$

   $\Leftrightarrow (1 + \cos x)\left(\dfrac{1 – \cos x}{1 + \sin x} -1\right)= 0$

   $\Leftrightarrow \left[\begin{array}{l}\cos x = -1\\1 – \cos x = 1 + \sin x\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}\cos x = -1\\\sin x + \cos x = 0\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}x = \pi + k2\pi\\\sin\left(x + \dfrac{\pi}{4}\right) = 0\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}x = \pi + k2\pi\\x =- \dfrac{\pi}{4}+ k\pi\end{array}\right. \quad (k \in \Bbb Z)$

   e) $4\cos^2x – 2\cos2x – 2\cos^22x = 0$

   $\Leftrightarrow 2(1+ \cos2x) – 2\cos2x – 2\cos^22x = 0$

   $\Leftrightarrow \cos^22x -1 = 0$

   $\Leftrightarrow 1 + \cos4x = 2$

   $\Leftrightarrow \cos4x = 1$

   $\Leftrightarrow 4x = k2\pi$

   $\Leftrightarrow x = k\dfrac{\pi}{2} \quad (k \in \Bbb Z)$

   Bình luận

2. Đáp án:

    

   Giải thích các bước giải:

    

   Bình luận