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

2 Answers
Jun 22, 2018

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

Jun 23, 2018

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