The double-angle formula for sine gives
[tex]\sin(2x)=2\sin x\cos x[/tex]
so
[tex]\sin(2x)+\cos x=2\sin x\cos x+\cos x=\cos x(2\sin x+1)=0[/tex]
Then either
[tex]\cos x=0\implies x=\dfrac\pi2,\dfrac{3\pi}2[/tex]
or
[tex]2\sin x+1=0\implies\sin x=-\dfrac12\implies x=\dfrac{7\pi}6,\dfrac{11\pi}6[/tex]
