Question

If x belongs to (0,π/2), the number of solutions of the equation sin7x+sin4x+sinx=0?

Answer 1

`x=pi/4`

Explanation:

Here,

`sin7x+sin4x+sinx=0 , where , x in(0,pi/2)`

`=>sin7x+sinx+sin4x=0`

`=>2sin((7x+x)/2)cos((7x-x)/2)+sin4x=0`

`=>2sin4xcos3x+sin4x=0`

`=>sin4x(2cos3x+1)=0`

`=>sin4x=0 or2cos3x=-1`

`=>sin4x=0 orcos3x=-1/2`

`(1)sin4x=0=>2sin2xcos2x=0`

`=>sin2x=0 or cos2x=0`

`=>2sinxcosx=0 or 2cos^2x-1=0`

`=>sinx=0 or cosx=0 orcos^2x=1/2`

`=>sinx=0 or cosx=0, cosx=-1/sqrt2 or cosx=1/sqrt2`

`color(red)((i)sinx=0=>x=0 !in(0,pi/2)`

`color(red)((ii)cosx=0=>x=pi/2!in(0,pi/2)`

`color(red)((iii)cosx=-1/sqrt2 < 0=>x!in(0,pi/2)`

`color(blue)((iv)cosx=1/sqrt2=>x=pi/4 in(0,pi/2)`

`(2)cos3x=0=>4cos^3x-3cosx=0`

`=>cosx(4cos^2x-3)=0`

`=>cosx=0 or 4cos^2x=3`

`=>cosx=0 or cos^2x=3/4=(sqrt3/2)^2`

`=>cosx=0 or cosx=-sqrt3/2 or cosx=sqrt3/2`

`color(red)((i)cosx=0=>x=pi/2 !in(0,pi/2)`

`color(red)((ii)cosx=-sqrt3/2 <0=>x !in (0,pi/2)`

`color(blue)((iii)cosx=sqrt3/2=>x=pi/6 in(0,pi/2)`

But ,

`x=pi/6=>sin7(pi/6)+sin4(pi/6)+sin(pi/6)!=0`

So, `color(red)(x!=pi/6`

Hence,

`x=pi/4`

Answer 2

3 solutions for [0, pi/2]
`0, (2pi)/9, pi/2`

Explanation:

sin 7x + sin x + sin 4x = 0
2sin 4x cos 3x + sin 4x = 0
sin 4x(2cos 3x + 1) = 0
Either factor should be zero
a. sin 4x = 0 -->
`4x = 2kpi` --> `x = (2kpi)/4 = (kpi)/2`
k = 0 --> x = 0; if k = 1 --> `x = pi/2`
b. 2cos 3x + 1 = 0 --> `cos 3x = -1/2`
Trig table and unit circle give:
`3x = +- (2pi)/3`
c. `3x = (2pi)/3 + 2kpi`
`x = (2pi)/9 + (2kpi)/3`
k = 0 --> `x = (2pi)/9`
k = 1 --> `x = (2pi)/9 + (2pi)/3 = (8pi)/9` (out of range)
d. `x = - (2pi)/3 + 2kpi` (out of range)
For `[0, pi/2]`, there are 3 solutions:
`0, (2pi)/9, pi/2`