The-number-of-solutions-of-the-equation-sin-5-1-sin-1-cos-cos-5-where-0-pi-2-is-

Question Number 18962 by Tinkutara last updated on 02/Aug/17

The number of solutions of the equation

\sin^{5} θ + \frac{1}{\sin θ} = \frac{1}{\cos θ} + \cos^{5} θ where

θ \in (0, \frac{π}{2}), is

Answered by behi.8.3.4.1.7@gmail.com last updated on 02/Aug/17

{s i n}^{6} θ . c o s θ + c o s θ = s i n θ + {c o s}^{6} θ . s i n θ

\Rightarrow s i n θ . c o s θ ({s i n}^{5} θ - {c o s}^{5} θ) - (s i n θ - c o s θ) = 0

s i n θ = a, c o s θ = b

\Rightarrow a b (a^{5} - b^{5}) - (a - b) = 0

\Rightarrow [a b (a^{4} + a^{3} b + a^{2} b^{2} + {a b}^{3} + b^{4}) - 1] (a - b) = 0

1) a = b \Rightarrow s i n θ = c o s θ \Rightarrow t g θ = 1 \Rightarrow θ = k π + \frac{π}{4}, k \in Z

2) a^{4} + b^{4} = {(a^{2} + b^{2})}^{2} - 2 a^{2} b^{2} = 1 - 2 a^{2} b^{2}

a^{3} b + {a b}^{3} = a b (a^{2} + b^{2}) = a b

\Rightarrow a b (1 - 2 a^{2} b^{2} + a^{2} b^{2} + a b) - 1 = 0

\Rightarrow a b (1 + a b - a^{2} b^{2}) - 1 = 0 (a b = t)

\Rightarrow t (1 + t - t^{2}) - 1 = 0 \Rightarrow - t^{3} + t^{2} + t - 1 = 0

\Rightarrow t^{3} - t^{2} - t + 1 = 0 \Rightarrow t^{2} (t - 1) - (t - 1) = 0

\Rightarrow (t - 1) (t^{2} - 1) = 0 \Rightarrow {(t - 1)}^{2} (t + 1) = 0

\Rightarrow t = 1, - 1 \Rightarrow s i n θ . c o s θ = \pm 1 \Rightarrow s i n 2 θ = \pm 2

t h i s e q . h a s n o t r e a l a n s w e r s .

Commented by Tinkutara last updated on 02/Aug/17

Thank you very much behi Sir!

Commented by behi.8.3.4.1.7@gmail.com last updated on 04/Aug/17

s i n 2 θ = 2 \Rightarrow \frac{e^{2 θ} - e^{- 2 θ}}{2 i} = 2 (e^{2 θ} = a)

\Rightarrow \frac{a - a^{- 1}}{2 i} = 2 \Rightarrow a^{2} - 4 a i - 1 = 0

\Rightarrow a = \frac{4 i \pm \sqrt{- 16 + 4}}{2} = \frac{4 i \pm 2 \sqrt{3} i}{2} = (2 \pm \sqrt{3}) i

\Rightarrow e^{2 θ} = (2 \pm \sqrt{3}) i \Rightarrow 2 θ = l n (2 \pm \sqrt{3}) + l n i

\Rightarrow θ = \frac{1}{2} l n (2 \pm \sqrt{3}) + \frac{i π}{4} (c o m p l e x a n s w e r) .

Commented by Tinkutara last updated on 04/Aug/17

But using this value in calculator does

not gives 2 . \sin (\ln (2 + \sqrt{3}) + \frac{i π}{2})

= 2.423 + 0.578 i