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 x=π4

Explanation:

Here,

sin7x+sin4x+sinx=0 , where , x in(0,pi/2)sin7x+sin4x+sinx=0,where,x(0,π2)

=>sin7x+sinx+sin4x=0sin7x+sinx+sin4x=0

=>2sin((7x+x)/2)cos((7x-x)/2)+sin4x=02sin(7x+x2)cos(7xx2)+sin4x=0

=>2sin4xcos3x+sin4x=02sin4xcos3x+sin4x=0

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

=>sin4x=0 or2cos3x=-1sin4x=0or2cos3x=1

=>sin4x=0 orcos3x=-1/2sin4x=0orcos3x=12

(1)sin4x=0=>2sin2xcos2x=0(1)sin4x=02sin2xcos2x=0

=>sin2x=0 or cos2x=0sin2x=0orcos2x=0

=>2sinxcosx=0 or 2cos^2x-1=02sinxcosx=0or2cos2x1=0

=>sinx=0 or cosx=0 orcos^2x=1/2sinx=0orcosx=0orcos2x=12

=>sinx=0 or cosx=0, cosx=-1/sqrt2 or cosx=1/sqrt2sinx=0orcosx=0,cosx=12orcosx=12

color(red)((i)sinx=0=>x=0 !in(0,pi/2)(i)sinx=0x=0(0,π2)

color(red)((ii)cosx=0=>x=pi/2!in(0,pi/2)(ii)cosx=0x=π2(0,π2)

color(red)((iii)cosx=-1/sqrt2 < 0=>x!in(0,pi/2)(iii)cosx=12<0x(0,π2)

color(blue)((iv)cosx=1/sqrt2=>x=pi/4 in(0,pi/2)(iv)cosx=12x=π4(0,π2)

(2)cos3x=0=>4cos^3x-3cosx=0(2)cos3x=04cos3x3cosx=0

=>cosx(4cos^2x-3)=0cosx(4cos2x3)=0

=>cosx=0 or 4cos^2x=3cosx=0or4cos2x=3

=>cosx=0 or cos^2x=3/4=(sqrt3/2)^2cosx=0orcos2x=34=(32)2

=>cosx=0 or cosx=-sqrt3/2 or cosx=sqrt3/2cosx=0orcosx=32orcosx=32

color(red)((i)cosx=0=>x=pi/2 !in(0,pi/2)(i)cosx=0x=π2(0,π2)

color(red)((ii)cosx=-sqrt3/2 <0=>x !in (0,pi/2)(ii)cosx=32<0x(0,π2)

color(blue)((iii)cosx=sqrt3/2=>x=pi/6 in(0,pi/2)(iii)cosx=32x=π6(0,π2)

But ,

x=pi/6=>sin7(pi/6)+sin4(pi/6)+sin(pi/6)!=0x=π6sin7(π6)+sin4(π6)+sin(π6)0

So, color(red)(x!=pi/6xπ6

Hence,

x=pi/4 x=π4

Jun 23, 2018

3 solutions for [0, pi/2]
0, (2pi)/9, pi/20,2π9,π2

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 = 2kpi4x=2kπ --> x = (2kpi)/4 = (kpi)/2x=2kπ4=kπ2
k = 0 --> x = 0; if k = 1 --> x = pi/2x=π2
b. 2cos 3x + 1 = 0 --> cos 3x = -1/2cos3x=12
Trig table and unit circle give:
3x = +- (2pi)/33x=±2π3
c. 3x = (2pi)/3 + 2kpi3x=2π3+2kπ
x = (2pi)/9 + (2kpi)/3x=2π9+2kπ3
k = 0 --> x = (2pi)/9x=2π9
k = 1 --> x = (2pi)/9 + (2pi)/3 = (8pi)/9x=2π9+2π3=8π9 (out of range)
d. x = - (2pi)/3 + 2kpix=2π3+2kπ (out of range)
For [0, pi/2][0,π2], there are 3 solutions:
0, (2pi)/9, pi/20,2π9,π2