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Question Number 17815 by Tinkutara last updated on 11/Jul/17

For x \in (0, π), the equation

\sin x + 2 \sin 2 x - \sin 3 x = 3 has

(1) Infinitely many solutions

(2) Three solutions

(3) One solution

(4) No solution

Commented by alex041103 last updated on 11/Jul/17

Commented by alex041103 last updated on 11/Jul/17

Clearly there are three solutions

x \approx 1.2094 and f (x) \approx 3

x \approx 1.3407 and f (x) \approx 3

x \approx 3.1239 and f (x) \approx 3

Commented by alex041103 last updated on 11/Jul/17

But for x = 0.1

f (x) = 0.1 + 2 \sin (0.2) - \sin (0.3) \approx 0.2018

For x = 3.13 (let say)

f (x) = 3.13 + 2 \sin (6.26) - \sin (9.39) \approx 3.0489

\Rightarrow f (0.1) < 3 < f (3.13) and f (x) is continious

\Rightarrow f (x) crosses y = 3 at least once where 0 < x < π

Commented by alex041103 last updated on 11/Jul/17

I^{'} m so sorry sir for misreading the question

Answered by b.e.h.i.8.3.417@gmail.com last updated on 11/Jul/17

x - \frac{x^{3}}{6} + 2 (2 x - \frac{8 x^{3}}{6}) - (3 x - \frac{27 x^{3}}{6}) = 3

2 x - \frac{1 + 16 - 27}{6} x^{3} = 3 \Rightarrow 2 x - \frac{5}{3} x^{3} = 3

\Rightarrow 5 x^{3} - 6 x + 9 = 0 \Rightarrow x = - 1.54 \notin (0, π)

\Rightarrow (4) n o s o l u t i o n .

n o t e : i t i s n o t a p e r f e c t w a y f o r t h i s Q .

b u t y o u c a n u s e t h i s m e t h o d f o r f i n d i n g

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

Commented by b.e.h.i.8.3.417@gmail.com last updated on 11/Jul/17

y e s y o u a r e r i g h t . i h a v e a m i s t a k e .

t h e c u b i c e q . h a s o n l y t h i s r e a l a n s w e r .

t h a t n o t l i e s i n : (0, π) . s o e q . h a s n o t

a n y s o l u t i o n i n : (0, π) \Rightarrow a n s w e r (4) .

Commented by alex041103 last updated on 11/Jul/17

You are using the first two terms of the taylor

expansion for \sin (x), {aren}^{'} t you

Commented by b.e.h.i.8.3.417@gmail.com last updated on 11/Jul/17

y e s, m r a l e x .