Find the general solution of the equation $\sin 2 x+\cos x=0$

Vedclass pdf generator app on play store
Vedclass iOS app on app store

$\sin 2 x+\cos x=0$

$\Rightarrow 2 \sin x \cos x+\cos x=0$

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

$\Rightarrow \cos x=0 \quad$ or

$2 \sin x+1=0$

Now, $\cos x=0 \Rightarrow \cos x=(2 n+1) \frac{\pi}{2},$ where $n \in Z$

$2 \sin x+1=0$

$\Rightarrow \sin x=\frac{-1}{2}=-\sin \frac{\pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right)=\sin \left(\pi+\frac{\pi}{6}\right)=\sin \frac{7 \pi}{6}$

$\Rightarrow x=n \pi+(-1)^{n} \frac{7 \pi}{6},$ where $n \in Z$

Therefore, the general solution is $(2 n+1) \frac{\pi}{2}$ or $n \pi+(-1)^{n} \frac{7 \pi}{6}, n \in Z$

Similar Questions

Let $S\, = \,\left\{ {\theta \, \in \,[ - \,2\,\pi ,\,\,2\,\pi ]\,  :\,2\,{{\cos }^2}\,\theta \, + \,3\,\sin \,\theta \, = \,0} \right\}$. Then the sum of the elements of $S$ is

  • [JEE MAIN 2019]

If $|cos\ x + sin\ x| + |cos\ x\ -\ sin\ x| = 2\ sin\ x$ ; $x \in  [0,2 \pi ]$ , then maximum integral value of $x$ is

General value of $\theta $ satisfying the equation ${\tan ^2}\theta + \sec 2\theta - = 1$ is

  • [IIT 1996]

If the equation $tan^4x -2sec^2x + [a]^2 = 0$ has atleast one solution, then the complete range of $'a'$ (where $a \in R$ ) is 
(Note : $[k]$ denotes greatest integer less than or equal to $k$ )

If $\theta $ and $\phi $ are acute satisfying $\sin \theta = \frac{1}{2},$ $\cos \phi = \frac{1}{3},$ then $\theta + \phi \in $

  • [IIT 2004]