It is given that
sin 2x + cos x = 0
We can write it as
2 sin x cos x + cos x = 0
cos x (2 sin x + 1) = 0
cos x = 0 or 2 sin x + 1 = 0
Let cos x = 0
Or, \( sinx=-\dfrac{1}{2} \)
Hence we have,
\( x=(2n+1)\dfrac{\pi}{2} \), where \( n \in Z\)
\( -sin=\dfrac{\pi}{6}\)
\( sin=(\pi-\dfrac{7\pi}{6})\)
\( sin=(\dfrac{7\pi}{6})\)
Which implies
\( x=nπ+(-1)^n\dfrac{7\pi}{6}\), where \( n \in Z\)
Therefore, the general solution is\( (2n +1)\dfrac{\pi}{2}\)
or \( nπ+(-1)^n\dfrac{7\pi}{6}, n \in Z\)
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