How-many-real-solutions-does-the-equation-x-sin3x-have-

Question Number 210702 by depressiveshrek last updated on 17/Aug/24

How many real solutions does the

equation x = \sin 3 x have ?

Answered by Frix last updated on 17/Aug/24

x = \sin 3 x

t = 3 x

\frac{t}{3} = \sin t

t_{1} = 0

- 1 ⩽ \frac{t}{3} ⩽ 1

- 3 ⩽ t ⩽ 3

\sin t ⩾ 0 \Rightarrow 0 ⩽ t ⩽ π; π > 3

\Rightarrow 1 solution for t \in (0, 3]

1 solution for t \in [- 3, 0)

\Rightarrow 3 solutions

\frac{t}{n} = \sin t

- n ⩽ t ⩽ n

\sin t ⩾ 0 \Rightarrow 2 k π ⩽ t ⩽ (2 k + 1) π

First time more than 3 solutions possible

at n = 7

- 7 ⩽ t ⩽ 7

0 ⩽ t ⩽ π \lor 2 π ⩽ t ⩽ 3 π

but in fact \frac{t}{7} > \sin t; 2 π ⩽ t

n = 8 \Rightarrow \frac{t}{8} = \sin t has 7 solutions

Answered by mr W last updated on 17/Aug/24

s a y 3 x = t, t h e e q u a t i o n b e c o m e s

\frac{t}{3} = \sin t . {l e t}^{'} s g e n e r a l l l o o k a t t h e

e q u a t i o n k x = \sin x .

w e s e e x = 0 i s a l w a y s a r o o t . b e s i d e s

d u e t o s y m m e t r y i f x = a i s a r o o t,

t h e n x = - a i s a l s o a r o o t . t h e r e f o r e

t h e n u m b e r o f r o o t s i s a l w a y s o d d .

d u e t o s y m m e t r y w e o n l y n e e d t o

t r e a t k ⩾ 0 . f o r x = 0, \sin x = 0 h a s

i n f i n i t e r o o t s . s o n o w w e t a k e k > 0 .

Commented by mr W last updated on 17/Aug/24

Commented by mr W last updated on 17/Aug/24

{(k x)}^{'} = {(\sin x)}^{'}

\Rightarrow k = \cos x = \sqrt{1 - k^{2} x^{2}} \Rightarrow x = \sqrt{\frac{1}{k^{2}} - 1}

2 (n - 1) π < x < (2 (n - 1) + \frac{1}{2}) π

4 {(n - 1)}^{2} π^{2} < x^{2} < 4 {(n - \frac{3}{4})}^{2} π^{2}

4 {(n - 1)}^{2} π^{2} + 1 < \frac{1}{k_{n}^{2}} < 4 {(n - \frac{3}{4})}^{2} π^{2} + 1

\frac{1}{\sqrt{4 {(n - \frac{3}{4})}^{2} π^{2} + 1}} < k_{n} < \frac{1}{\sqrt{4 {(n - 1)}^{2} π^{2} + 1}}

k_{n} = \cos \sqrt{\frac{1}{k_{n}^{2}} - 1}

t h i s e q u a t i o n h a s i n f i n i t e s o l u t i o n s :

k_{1}, k_{2}, k_{3}, \dots

k_{1} = 1

k_{2} \approx \frac{1}{7.789706}

k_{3} \approx \frac{1}{14.101695}

k_{4} \approx \frac{1}{20.395833}

\dots \dots

a t x = 0 : k = k_{1} = 1 .

i f k ⩾ k_{1} = 1, t h e r e i s o n e r o o t x = 0 .

i f k < k_{1} = 1, t h e r e a r e a t l e a s t 3 r o o t s .

g e n e r a l l y

f o r k = k_{n} t h e r e a r e 4 n - 3 r o o t s

f o r k_{n + 1} < k < k_{n} t h e r e a r e 4 n - 1 r o o t s

b a c k t o t h e o r i g i n a l q u e s t i o n w i t h

x = \sin 3 x .

i t i s t h e s a m e a s \frac{x}{3} = \sin x

s i n c e k = \frac{1}{3} < k_{1}, > k_{2} \Rightarrow i t h a s 3 r o o t s .

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