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

Jun 22, 2018

$x = \frac{\pi}{4}$

#### Explanation:

Here,

$\sin 7 x + \sin 4 x + \sin x = 0 , w h e r e , x \in \left(0 , \frac{\pi}{2}\right)$

$\implies \sin 7 x + \sin x + \sin 4 x = 0$

$\implies 2 \sin \left(\frac{7 x + x}{2}\right) \cos \left(\frac{7 x - x}{2}\right) + \sin 4 x = 0$

$\implies 2 \sin 4 x \cos 3 x + \sin 4 x = 0$

$\implies \sin 4 x \left(2 \cos 3 x + 1\right) = 0$

$\implies \sin 4 x = 0 \mathmr{and} 2 \cos 3 x = - 1$

$\implies \sin 4 x = 0 \mathmr{and} \cos 3 x = - \frac{1}{2}$

$\left(1\right) \sin 4 x = 0 \implies 2 \sin 2 x \cos 2 x = 0$

$\implies \sin 2 x = 0 \mathmr{and} \cos 2 x = 0$

$\implies 2 \sin x \cos x = 0 \mathmr{and} 2 {\cos}^{2} x - 1 = 0$

$\implies \sin x = 0 \mathmr{and} \cos x = 0 \mathmr{and} {\cos}^{2} x = \frac{1}{2}$

$\implies \sin x = 0 \mathmr{and} \cos x = 0 , \cos x = - \frac{1}{\sqrt{2}} \mathmr{and} \cos x = \frac{1}{\sqrt{2}}$

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)

$\left(2\right) \cos 3 x = 0 \implies 4 {\cos}^{3} x - 3 \cos x = 0$

$\implies \cos x \left(4 {\cos}^{2} x - 3\right) = 0$

$\implies \cos x = 0 \mathmr{and} 4 {\cos}^{2} x = 3$

$\implies \cos x = 0 \mathmr{and} {\cos}^{2} x = \frac{3}{4} = {\left(\frac{\sqrt{3}}{2}\right)}^{2}$

$\implies \cos x = 0 \mathmr{and} \cos x = - \frac{\sqrt{3}}{2} \mathmr{and} \cos x = \frac{\sqrt{3}}{2}$

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 ,

$x = \frac{\pi}{6} \implies \sin 7 \left(\frac{\pi}{6}\right) + \sin 4 \left(\frac{\pi}{6}\right) + \sin \left(\frac{\pi}{6}\right) \ne 0$

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

Hence,

$x = \frac{\pi}{4}$

Jun 23, 2018

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

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