Knowing that \(\sec x=\dfrac{1}{\cos x}\):
\(\sec\theta=4\iff\dfrac{1}{\cos\theta}=4\iff\cos\theta=\dfrac{1}{4}\)
Using that \(\sin^2x+\cos^2x=1\):
\(\sin^2\theta+(\dfrac{1}{4})^2=1\iff \sin^2\theta=\dfrac{15}{16}\iff\\\\\boxed{\sin\theta=\pm\dfrac{\sqrt{15}}{4}}\)
