How do you solve sin(2x) - sin(3x) + sin(4x) = 0?

1 Answer
May 11, 2015

The answer is {x in RR : x=kpi or x=(2pi)/3+2kpi or x=-(2pi)/3+2kpi, k in ZZ}

Remember:
sin(4x)=2sin(2x)cos(2x)
sin(3x)=sin(x)cos(2x)+cos(x)sin(2x)

sin(2x)(1+2cos(2x))=sin(2x)cos(x)+cos(2x)sin(x)

Remember: 1=2-1

sin(2x)(2+2cos(2x)-1-cos(x))=cos(2x)sin(x)

Remember: 1+cos(2x)=2cos^2(x)

2sin(x)cos(x)(4cos^2(x)-1-cos(x))=cos(2x)sin(x)

We have one set of solutions: A={x in RR :sin(x)=0}={kpi}_(k in ZZ)

We now can simplify

2cos(x)(4cos^2(x)-1-cos(x))=cos(2x)
Remember: cos(2x)=cos^2(x)-sin^2(x)=2cos^2(x)-1
8cos^3(x)-2cos(x)-2cos^2(x)=2cos^2(x)-1

y=cos(x)

8y^3-4y^2-2y+1=0

We notice it's

(2y)^3-(2y)^2-(2y)+1=0
So 2y=-1 and it's the sole solution in RR (we can divide with Ruffini's rule, or calculate the local minimum to prove it)

So we have another set of solutions B={x in RR : cos(x)=-1/2}={(2pi)/3+2kpi} cup {-(2pi)/3+2kpi}

We know we don't have any other solution so the set is A cup B