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#