Тригонометрия. сколько корней имеет уравнение соs2x+sin6x=cos2x на отрезке -2π, 2π

Sin6x=0
6x=πn,n∈z
x=πn/6,n∈z
-2π≤πn/6≤2π
-12≤n≤12
n=-12⇒x=-2π
n=-11⇒x=-11π/6
n=-10⇒x=-5π/3
n=-9⇒x=-3π/2
n=-8⇒x=-4π/3
n=-7⇒x=-7π/6
n=-6⇒x=-π
n=-5⇒x=-5π/6
n=-4⇒x=-2π/3
n=-3⇒x=-π/2
n=-2⇒x=-π/3
n=-1⇒x=-π/6
n=0⇒x=0
n=12⇒x=2π
n=11⇒x=11π/6
n=10⇒x=5π/3
n=-9⇒x=-3π/2
n=8⇒x=4π/3
n=7⇒x=7π/6
n=6⇒x=π
n=5⇒x=5π/6
n=4⇒x=2π/3
n=3⇒x=π/2
n=2⇒x=π/3
n=1⇒x=π/6
25 корней