4sin-2-x-4sin-x-cos-2-x-2cos-x-0-lt-x-lt-pi-2-Find-sin-x-

Question Number 173787 by mnjuly1970 last updated on 18/Jul/22

4 {s i n}^{2} (x) - 4 s i n (x) = {c o s}^{2} (x) - 2 c o s (x)

, 0 < x < \frac{π}{2} . F i n d s i n (x) = ?

Commented by infinityaction last updated on 18/Jul/22

\frac{1}{\sqrt{5}}

Commented by som(math1967) last updated on 18/Jul/22

\frac{3}{5} a l s o

Commented by infinityaction last updated on 18/Jul/22

{4 \sin}^{2} x - 4 \sin x + 1 = \cos^{2} x - 2 \cos x + 1

{(1 - 2 \sin x)}^{2} = {(1 - \cos x)}^{2}

{(1 - 2 \sin x)}^{2} - {(1 - \cos x)}^{2} = 0

(\cos x - 2 \sin x) (2 - 2 \sin x - \cos x) = 0

i f \cos x - 2 \sin x = 0 \Rightarrow \tan x = \frac{1}{2}

\sin x = \frac{1}{\sqrt{5}}

2 - 2 \sin x = \sqrt{1 - \sin^{2} x}

{5 \sin}^{2} x - 8 \sin x + 3 = 0

(\sin x - 1) (5 \sin x - 3) = 0

\sin x = 1 (r e j e c t e d)

\sin x = \frac{3}{5}

Commented by mnjuly1970 last updated on 18/Jul/22

t h a n k y o u s o m u c h s i r . .

Commented by MJS_new last updated on 18/Jul/22

why do you reject 1 ?

Commented by infinityaction last updated on 18/Jul/22

g i v e n i n t h e q u e s t i o n

0 < x < \frac{π}{2}

Commented by Tawa11 last updated on 18/Jul/22

Great sirs

Commented by MJS_new last updated on 18/Jul/22

ok {didn}^{'} t notice \dots

Commented by infinityaction last updated on 18/Jul/22

n o p r o b l e m s i r y o u r a l g e b r a

i s v e r y g o o d

i a m s t u d e n t o f u n d e r g r u a t e

s o c a n y o u s u g g e s t m e a n

a l g e b r a b o o k ? ?

Commented by MJS_new last updated on 18/Jul/22

sorry I {can}^{'} t . I studied math more than 20

years ago but I became a musician . math is

only my recreational sport now \dots

Commented by infinityaction last updated on 19/Jul/22

o k s i r t h a n k s

Answered by MJS_new last updated on 18/Jul/22

x = 2 \arctan t

\frac{- 8 t^{3} + 16 t^{2} - 8 t}{{(t^{2} + 1)}^{2}} = \frac{3 t^{4} - 2 t^{2} - 1}{{(t^{2} + 1)}^{2}}

- 8 t {(t - 1)}^{2} = (t - 1) (t + 1) (3 t^{2} + 1)

t_{1} = 1

3 t^{3} + 11 t^{2} - 7 t + 1 = 0

(3 t - 1) (t + 2 - \sqrt{5}) (t + 2 + \sqrt{5}) = 0

t_{2} = \frac{1}{3}

t_{3} = - 2 + \sqrt{5}

t_{4} = - 2 - \sqrt{5}

\sin x = \sin 2 \arctan t = \frac{2 t}{t^{2} + 1}

\sin x \in {- \frac{\sqrt{5}}{5}, \frac{\sqrt{5}}{5}, \frac{3}{5}, 1}

Answered by a.lgnaoui last updated on 18/Jul/22

4 \sin^{2} (x) - 4 \sin (x) = 1 - \sin^{2} (x) - 2 \sqrt{1 - \sin^{2} (x)}

p o s i n s \sin (x) = z

5 z^{2} - 4 z - 1 = - 2 \sqrt{1 - z^{2}}

{[z - 1]}^{2} {[5 z + 1]}^{2} = 4 (1 - z^{2})

(1 - z) {(5 z + 1)}^{2} = 4 (1 + z)

(1 - z) (25 z^{2} + 10 z + 1) = 4 z + 4

25 z^{2} + 10 z + 1 - 25 z^{3} - 10 z^{2} - z - 4 z - 4 = 0

- 25 z^{3} + 15 z^{2} - 5 z - 3 = 0

25 z^{3} - 15 z^{2} + 5 z + 3 = 0

z^{3} - \frac{3}{5} z^{2} + \frac{z}{5} + \frac{3}{25} = 0

Z = z + 1 / 5

Z^{3} = z^{3} + \frac{3}{5} z^{2} + \frac{3}{25} z + \frac{1}{125}

- \frac{3}{5} Z^{2} = \frac{- 3}{5} z^{2} + - \frac{6}{25} z + \frac{- 3}{125}

Z = z + \frac{1}{5}

Z^{3} + (\frac{2}{25}) Z + \frac{22}{125} = 0 p = \frac{2}{25} q = \frac{22}{125}

p^{3} = (4 \times 8 / 5^{6}) / 27 + (4 \times 11^{2}) /

Δ = 484 / 5^{6} + \frac{32}{27 \times 5^{6}} = \frac{484 \times 27 + 32}{3^{3} \times 5^{6}} = \frac{13068 + 32}{} = \frac{13100}{421875} = 0, 031 \Rightarrow \sqrt{Δ} = 0, 176

q = \frac{22}{125} = 0, 176 \Rightarrow \frac{q}{2} = 0, 088

= - q / 2 \pm [^{3} \sqrt{\frac{- q}{2} \pm (4 p^{3} + 27 q^{2}) / 27)}] / 2 \sqrt{0, 176 - 0, 088}

Z = 0, 088 + [^{3} \sqrt{0, 088 + 0, 176}] / 2

Z = 0, 088 + [^{3} \sqrt{- 0, 088 + 0, 176}] / 2

Z = 0, 64 e t Z = - 0, 56

z = Z - \frac{1}{5} \Rightarrow \sin (x) = 0, 44

\sin (x) = 0, 36