Решите уравнение sin x sin 3x = 1/2

Решите задачу:

$sinx*sin3x=\dfrac{1}{2}\\\\Formyla\;:\\\\\boxed{\;\;sin\alpha*sin\beta=\dfrac{1}{2}*\Big(\;cos\big(\alpha-\beta\big)-cos\big(\alpha+\beta\big)\;\Big)\;\;}\\\\\\\boxed{\big{\;\;cos2x=2cos^2x-1\;\;}}\\\\\\\dfrac{1}{2}*\Big(cos2x-cos4x\Big)=\dfrac{1}{2}\\\\cos2x-cos4x=1\\\\cos2x-\Big(2cos^22x-1\Big)=1\\\\cos2x-2cos^22x+1=1\\\\2cos^22x-cos2x=0\\\\cos2x*\Big(\;2cos2x-1\;\Big)=0\\\\cos2x*\Big(\;cos2x-\dfrac{1}{2}\;\Big)=0\\\\$

$1)\;\;\;cos2x=0\\\\2x=\dfrac{\pi}{2}+\pi\;n\;\;,\;\;\;n\;\in\;\mathbb{Z}\\\\\boxed{\;\;x\;_k=\dfrac{\pi}{4}+\frac{\pi\;k}{2}\;\;,\;\;\;n\;\in\;\mathbb{Z}\;\;}\\\\\\2)\;\;\;cos2x-\dfrac{1}{2}=0\\\\cos2x=\dfrac{1}{2}\\\\2x=\pm\;\dfrac{\pi}{3}+2\pi\;p\;\;,\;\;\;p\;\in\;\mathbb{Z}\\\\\boxed{\Big{\;\;x\;_m=\pm\;\dfrac{\pi}{6}+\pi\;m\;\;,\;\;\;m\;\in\;\mathbb{Z}\;\;}}\\$