Question

Question #6527d

Answer 1

`x=pi/4,(7pi)/12,(11pi)/12.`

Explanation:

`sin3x+cos3x=0`
`rArrcos(pi/2-3x)+cos3x=0`
`rArr2cos((pi/2-3x+3x)/2)cos((pi/2-3x-3x)/2)=0`
`rArr2cos(pi/4)cos(pi/4-3x)=0`
`rArrcos(pi/4-3x)=0,where, 2cos(pi/4)!=0`
`rArrcos(3x-pi/4)=0`, as `cos(-theta)=costheta`
`rArr3x-pi/4=(2k+1)pi/2,kinZ,`
`rArr3x=(2k+1)pi/2+pi/4`
`color(red)(rArrx=(2k+1)pi/6+pi/12,kinZ)`,
So, taking `k=0,+-1.+-2,+-3...`
`k=0rArrx=pi/6+pi/12=color(blue)(pi/4in(0.pi),`
`k=1rArrx=(3pi)/6+pi/12=color(blue)((7pi)/12in(0.pi))`
# k=2rArrx=(5pi)/6+pi/12=color(blue)((11pi)/12in(0.pi),#
# k=3rArrx=(7pi)/6+pi/12=(15pi)/12!in(0.pi)#
Also,`k=-1rArrx=(-pi)/6+pi/12=-pi/12!in(0.pi)`
Hence, `x=pi/4,(7pi)/12,(11pi)/12`

Answer 2

`x=pi/4,(7pi)/12,(11pi)/12`
We know that,
`color(blue)((I) sin3theta=3sintheta-4sin^3theta)`
`color(blue)((II)cos3theta=4cos^3theta-3costheta)`
`color(violet)((III)sin2theta=2sinthetacostheta`

Explanation:

Here,

`color(blue)(sin3x+cos3x=0) ,where,color(red)(x in (0,pi)`

Using above `color(blue)((I)and (II)`

`color(blue)(3sinx-4sin^3x+4cos^3x-3cosx=0`

`=>4cos^3x-4sin^3x-3cosx+3sinx=0`

`=>4(cos^3x-sin^3x)-3(cosx-sinx)=0`

`=>4(cosx-sinx)(cos^2x+cosxsinx+sin^2x)-3(cosx-sinx)=0`

`=>(cosx-sinx)[4(cos^2x+cosxsinx+sin^2x)-3]=0`

`=>(cosx-sinx)[4(1+cosxsinx)-3]=0`

`=>(cosx-sinx)[4+4sinxcosx-3]=0`

`=>(cosx-sinx)(1+2*2sinxcosx)=0`

`=>(cosx-sinx)(1+2color(violet)(sin2x))=0...tocolor(violet)(Apply(III)`

`=>cosx-sinx=0 or 1+2sin2x=0=>sin2x=-1/2`

Now, `x in (0,pi)=>I^(st)Quadrant or II^(nd)Quadrant`

`(i)cosx-sinx=0=>sinx=cosx=>tanx=1 > 0`

`i.e.I^(st)Quadrant=>color(red)(x=pi/4`

Also, `x in(0,pi)`

`=>2x in(0,2pi)=>I^(st),II^(nd),III ^(rd),IV^(th)Quadrant`

`(ii)sin2x=-1/2 < 0=>III^(rd) or IV^(th) Quadrant`

`:.2x=pi+pi/6=(7pi)/6 or 2x=2pi-pi/6=(11pi)/6`

`i.e. color(red)( x=(7pi)/12 or x=(11pi)/12`

Hence, `x=pi/4,(7pi)/12,(11pi)/12`