Question

For 0< x<π/2, if `sinx+sin3x=cosx+cos3x` then x = ?

For `0<x<π/2` if `sinx+sin3x=cosx+cos3x` then `x=`?

Answer 1

`sinx+sin3x=cosx+cos3x`

`=2sin2xcosx=2cos2xcosx`

`=2sin2xcosx-2cos2xcosx=0`

`=2cosx(sin2x-cos2x)=0`

So for `cosx = 0` there is no solution as `0 < x < pi/2`

when `sin2x-cos2x=0`

`=>tan2x=1=tan(pi/4)`

`=>x=pi/8`

Answer 2

The solutions are `S={pi/2, pi/8}`

Explanation:

We need

`sina+sinb=2sin((a+b)/2)cos((a-b)/2)`

`cosa+cosb=2cos((a+b)/2)cos((a-b)/2)`

Therefore,

`sinx+sin3x=2sin((x+3x)/2)cos((3x-x)/2)=2sin2xcosx`

and

`cosx+cos3x=2cos((x+3x)/2)cos((3x-x)/2)=2cos2xcosx`

Therefore,

`sinx+sin3x=cosx+cos3x`

`2sin2xcosx=2cos2xcosx`

`2sin2xcosx-2cos2xcosx=0`

`2cosx(sin2x-cos2x)=0`

So,

`cosx=0`, `=>`, `x=pi/2+2kpi` and `x=3/2pi+2kpi`, `AA k in ZZ`

`sin2x-cos2x=0`, `=>`, `tan2x=1`, `=>`, `2x=pi/4+kpi`

`x=pi/8+k/2pi`, `AA k in ZZ`