Question

Find the general solution of the following equation:

\[\sin 2x + \cos x = 0\]

Answer 1

We have:

\[\sin2x + \cos x = 0\]

\[\Rightarrow \cos x = - \sin 2x\]

\[ \Rightarrow \cos x = \cos \left( \frac{\pi}{2} + 2x \right)\]

\[ \Rightarrow x = 2n\pi \pm \left( \frac{\pi}{2} + 2x \right), n \in Z\]

On taking positive sign, we have:

\[x = 2n\pi + \frac{\pi}{2} + 2x\]
\[ \Rightarrow - x = 2n\pi + \frac{\pi}{2}\]
\[ \Rightarrow x = 2m\pi - \frac{\pi}{2}, m = - n \in Z\]
\[ \Rightarrow x = \frac{(4m - 1)\pi}{2}, m \in Z\]

On taking negative sign, we have:

`x-2nx-x/2-2x`

`=>3x=2nx-pi/2`

`=>x=((4n-1)x)/6,n in "Z"`