Answer:
[tex]x=k\frac\pi2; k\in \mathbb{Z}[/tex]
Step-by-step explanation:
Let's use the formulas that change a sum of two sines into a product:
[tex]sin\ p + sin\ q =2 sin{\frac{p+q}2}cos{\frac{p-q}2}[/tex]
[tex]sin 3x+sinx =2sin \frac{3x+x}2 cos \frac{3x-x}2 = 2(sin2x)(cosx)=0[/tex]
At this point it's either [tex]sin 2x =0 \implies 2x= k\pi \rightarrow x = k\frac{\pi}2[/tex]
or [tex]cosx=0 \rightarrow x=\frac\pi2+ k\pi = (2k+1)\frac\pi2[/tex]
with the usual condition of [tex]k\in \mathbb{Z}[/tex].
Now if you look at both sets of solutions, first one considers all multiples of [tex]\frac\pi2[/tex], while the second only the odd multiples. it means that the first includes the second.
