15*sin(12*x)-10*sin(7*x)-10*sin(5*x)=0
asked by guest
on Mar 30, 2025 at 12:57 am
You asked:
Solve the equation
15sin(12x)−10sin(7x)−10sin(5x)=0 for the variable
x.
MathBot Answer:
x∈{2nπ∣n∈Z}∪{π(2n+1)∣n∈Z}∪{2π(4n+3)n∈Z}∪{2π(4n+1)n∈Z}∪{32π(3n+1)n∈Z}∪{32π(3n+2)n∈Z}∪{3π(6n+5)n∈Z}∪{3π(6n+1)n∈Z}∪{6π(12n+11)n∈Z}∪{6π(12n+5)n∈Z}∪{6π(12n+7)n∈Z}∪{6π(12n+1)n∈Z}∪{2nπ+2arctan(CRootOf(5x12−190x10+1495x8−2772x6+1475x4−206x2+1,6))n∈Z}∪{2nπ+2arctan(CRootOf(5x12−190x10+1495x8−2772x6+1475x4−206x2+1,7))n∈Z}∪{2nπ+2arctan(CRootOf(5x12−190x10+1495x8−2772x6+1475x4−206x2+1,8))n∈Z}∪{2nπ+2arctan(CRootOf(5x12−190x10+1495x8−2772x6+1475x4−206x2+1,9))n∈Z}∪{2nπ+2arctan(CRootOf(5x12−190x10+1495x8−2772x6+1475x4−206x2+1,10))n∈Z}∪{2nπ+2arctan(CRootOf(5x12−190x10+1495x8−2772x6+1475x4−206x2+1,11))n∈Z}∪{2nπ+2arctan(CRootOf(5x12−190x10+1495x8−2772x6+1475x4−206x2+1,0))+2πn∈Z}∪{2nπ+2arctan(CRootOf(5x12−190x10+1495x8−2772x6+1475x4−206x2+1,1))+2πn∈Z}∪{2nπ+2arctan(CRootOf(5x12−190x10+1495x8−2772x6+1475x4−206x2+1,2))+2πn∈Z}∪{2nπ+2arctan(CRootOf(5x12−190x10+1495x8−2772x6+1475x4−206x2+1,3))+2πn∈Z}∪{2nπ+2arctan(CRootOf(5x12−190x10+1495x8−2772x6+1475x4−206x2+1,4))+2πn∈Z}∪{2nπ+2arctan(CRootOf(5x12−190x10+1495x8−2772x6+1475x4−206x2+1,5))+2πn∈Z}