Question

Solve cos 3theta+cos 2theta=sin 3theta/2+sin theta/2?

`Solve cos 3theta+cos 2theta=sin (3theta/2)+sin (theta/2)?`

Answer 1

`cos3theta+cos2theta=sin((3theta)/2)+sin(theta/2)`

`=>2cos((5theta)/2)cos(theta/2)=2sin(theta)cos(theta/2)`

`=>2cos((5theta)/2)cos(theta/2)-2sin(theta)cos(theta/2)=0`

`=>2cos(theta/2)(cos((5theta)/2)-sin(theta))=0`

When

`cos(theta/2)=0`

`=>theta/2=(2n+1)pi/2" where " n in ZZ`

`=>theta=(2n+1)pi" where " n in ZZ`

When

`(cos((5theta)/2)-sin(theta))=0`

`=>cos((5theta)/2)=sin(theta)=cos(pi/2-theta)`

So

`(5theta)/2=2kpipm(pi/2-theta)"where "k in ZZ`

So

`(7theta)/2=2kpi+pi/2"where "k in ZZ`

`=>theta=4/7kpi+pi/7"where "k in ZZ`

And also

`(5theta)/2=2kpi-(pi/2-theta)"where "k in ZZ`

`=>(3theta)/2=2kpi-pi/2"where "k in ZZ`

`=>theta=4/3kpi-pi/3"where "k in ZZ`