text: 0 like 0 dislike
  14 views
  Solve $\sin 2 x+\cos 2 x=0$
  solved
  | 14 views
  
  0 like 0 dislike
  $\sin 2 x+\cos 2 x=0$
  $\sin 2 x=-\cos 2 x$
  $\frac{\sin 2 x}{\cos 2 x}=-1$
  $\tan 2 x=-1 \longrightarrow 1$
  $\pi<x<2 \pi$
  $2 \pi<2 x<4 \pi$
  $\tan x$ is negative in second and fourth quadrant
  $\therefore \tan \frac{3 \pi}{4}=\tan \frac{7 \pi}{4}=\ldots \ldots=\tan (4 n-1) \frac{\pi}{4}=-1$
  $n=1,2,3, \ldots$
  But $2 \pi<2 x<4 \pi$
  $\therefore 2 x=\frac{11 \pi}{4}$ or $2 x=\frac{15 \pi}{4}$
  $x=\frac{11 \pi}{8}$ or $\frac{15 \pi}{8}$
  by Diamond (38,951 points)
  
  0 like 0 dislike
  0 like 0 dislike
  0 like 0 dislike
  0 like 0 dislike
  0 like 0 dislike
  0 like 0 dislike
  0 like 0 dislike
  0 like 0 dislike
  0 like 0 dislike