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

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

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Answer 1 · 2018-06-22T19:43:19

$x = \frac{π}{4}$ $x=pi/4$

Explanation:

Here,

$sin 7 x + sin 4 x + sin x = 0, w h e r e, x \in (0, \frac{π}{2})$ $sin7x+sin4x+sinx=0 , where , x in(0,pi/2)$

$\Rightarrow sin 7 x + sin x + sin 4 x = 0$ $=>sin7x+sinx+sin4x=0$

$\Rightarrow 2 sin (\frac{7 x + x}{2}) cos (\frac{7 x - x}{2}) + sin 4 x = 0$ $=>2sin((7x+x)/2)cos((7x-x)/2)+sin4x=0$

$\Rightarrow 2 sin 4 x cos 3 x + sin 4 x = 0$ $=>2sin4xcos3x+sin4x=0$

$\Rightarrow sin 4 x (2 cos 3 x + 1) = 0$ $=>sin4x(2cos3x+1)=0$

$\Rightarrow sin 4 x = 0 or 2 cos 3 x = - 1$ $=>sin4x=0 or2cos3x=-1$

$\Rightarrow sin 4 x = 0 or cos 3 x = - \frac{1}{2}$ $=>sin4x=0 orcos3x=-1/2$

$(1) sin 4 x = 0 \Rightarrow 2 sin 2 x cos 2 x = 0$ $(1)sin4x=0=>2sin2xcos2x=0$

$\Rightarrow sin 2 x = 0 or cos 2 x = 0$ $=>sin2x=0 or cos2x=0$

$\Rightarrow 2 sin x cos x = 0 or 2 {cos}^{2} x - 1 = 0$ $=>2sinxcosx=0 or 2cos^2x-1=0$

$\Rightarrow sin x = 0 or cos x = 0 or {cos}^{2} x = \frac{1}{2}$ $=>sinx=0 or cosx=0 orcos^2x=1/2$

$\Rightarrow sin x = 0 or cos x = 0, cos x = - \frac{1}{\sqrt{2}} or cos x = \frac{1}{\sqrt{2}}$ $=>sinx=0 or cosx=0, cosx=-1/sqrt2 or cosx=1/sqrt2$

$(i) sin x = 0 \Rightarrow x = 0 \notin (0, \frac{π}{2})$ $color(red)((i)sinx=0=>x=0 !in(0,pi/2)$

$(i i) cos x = 0 \Rightarrow x = \frac{π}{2} \notin (0, \frac{π}{2})$ $color(red)((ii)cosx=0=>x=pi/2!in(0,pi/2)$

$(i i i) cos x = - \frac{1}{\sqrt{2}} < 0 \Rightarrow x \notin (0, \frac{π}{2})$ $color(red)((iii)cosx=-1/sqrt2 < 0=>x!in(0,pi/2)$

$(i v) cos x = \frac{1}{\sqrt{2}} \Rightarrow x = \frac{π}{4} \in (0, \frac{π}{2})$ $color(blue)((iv)cosx=1/sqrt2=>x=pi/4 in(0,pi/2)$

$(2) cos 3 x = 0 \Rightarrow 4 {cos}^{3} x - 3 cos x = 0$ $(2)cos3x=0=>4cos^3x-3cosx=0$

$\Rightarrow cos x (4 {cos}^{2} x - 3) = 0$ $=>cosx(4cos^2x-3)=0$

$\Rightarrow cos x = 0 or 4 {cos}^{2} x = 3$ $=>cosx=0 or 4cos^2x=3$

$\Rightarrow cos x = 0 or {cos}^{2} x = \frac{3}{4} = {(\frac{\sqrt{3}}{2})}^{2}$ $=>cosx=0 or cos^2x=3/4=(sqrt3/2)^2$

$\Rightarrow cos x = 0 or cos x = - \frac{\sqrt{3}}{2} or cos x = \frac{\sqrt{3}}{2}$ $=>cosx=0 or cosx=-sqrt3/2 or cosx=sqrt3/2$

$(i) cos x = 0 \Rightarrow x = \frac{π}{2} \notin (0, \frac{π}{2})$ $color(red)((i)cosx=0=>x=pi/2 !in(0,pi/2)$

$(i i) cos x = - \frac{\sqrt{3}}{2} < 0 \Rightarrow x \notin (0, \frac{π}{2})$ $color(red)((ii)cosx=-sqrt3/2 <0=>x !in (0,pi/2)$

$(i i i) cos x = \frac{\sqrt{3}}{2} \Rightarrow x = \frac{π}{6} \in (0, \frac{π}{2})$ $color(blue)((iii)cosx=sqrt3/2=>x=pi/6 in(0,pi/2)$

But ,

$x = \frac{π}{6} \Rightarrow sin 7 (\frac{π}{6}) + sin 4 (\frac{π}{6}) + sin (\frac{π}{6}) \neq 0$ $x=pi/6=>sin7(pi/6)+sin4(pi/6)+sin(pi/6)!=0$

So, $x \neq \frac{π}{6}$ $color(red)(x!=pi/6$

Hence,

$x = \frac{π}{4}$ $x=pi/4$

Answer 2 · 2018-06-23T04:17:37

3 solutions for [0, pi/2]
$0, \frac{2 π}{9}, \frac{π}{2}$ $0, (2pi)/9, pi/2$

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 -->
$4 x = 2 k π$ $4x = 2kpi$ --> $x = \frac{2 k π}{4} = \frac{k π}{2}$ $x = (2kpi)/4 = (kpi)/2$
k = 0 --> x = 0; if k = 1 --> $x = \frac{π}{2}$ $x = pi/2$
b. 2cos 3x + 1 = 0 --> $cos 3 x = - \frac{1}{2}$ $cos 3x = -1/2$
Trig table and unit circle give:
$3 x = \pm \frac{2 π}{3}$ $3x = +- (2pi)/3$
c. $3 x = \frac{2 π}{3} + 2 k π$ $3x = (2pi)/3 + 2kpi$
$x = \frac{2 π}{9} + \frac{2 k π}{3}$ $x = (2pi)/9 + (2kpi)/3$
k = 0 --> $x = \frac{2 π}{9}$ $x = (2pi)/9$
k = 1 --> $x = \frac{2 π}{9} + \frac{2 π}{3} = \frac{8 π}{9}$ $x = (2pi)/9 + (2pi)/3 = (8pi)/9$ (out of range)
d. $x = - \frac{2 π}{3} + 2 k π$ $x = - (2pi)/3 + 2kpi$ (out of range)
For $[0, \frac{π}{2}]$ $[0, pi/2]$ , there are 3 solutions:
$0, \frac{2 π}{9}, \frac{π}{2}$ $0, (2pi)/9, pi/2$

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

2 Answers

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