cos-x-sin-x-4-5-5sin-x-

Question Number 89047 by jagoll last updated on 15/Apr/20

\cos x + \sin x = \frac{4}{5}

5 \sin x = ?

Commented by john santu last updated on 15/Apr/20

\pm \sqrt{1 - \sin^{2} x} = \frac{4}{5} - \sin x

1 - \sin^{2} x = \frac{16}{25} - \frac{8}{5} \sin x + \sin^{2} x

2 \sin^{2} x - \frac{8}{5} \sin x - \frac{9}{25} = 0

\Rightarrow \sin x = \frac{(\frac{8}{5} - \frac{4 \sqrt{34}}{5})}{4}

5 \sin x = 2 - \sqrt{34} \approx - 3.831

Answered by MJS last updated on 15/Apr/20

t = \tan \frac{x}{2} \Leftrightarrow x = 2 \arctan t

\frac{1 - t^{2}}{t^{2} + 1} + \frac{2 t}{t^{2} + 1} = \frac{4}{5}

- 5 t^{2} + 10 t + 5 = 4 t^{2} + 4

t^{2} - \frac{10}{9} t - \frac{1}{9} = 0

t = \frac{5}{9} \pm \frac{\sqrt{34}}{9}

\Rightarrow 5 \sin x = \frac{10 t}{t^{2} + 1} = 2 \pm \frac{\sqrt{34}}{2}

Commented by john santu last updated on 15/Apr/20

t^{2} = \frac{59 \pm 10 \sqrt{34}}{81} \Rightarrow t^{2} + 1 = \frac{140 \pm 10 \sqrt{34}}{81}

10 t = \frac{50 \pm 10 \sqrt{34}}{9}

\frac{10 t}{t^{2} + 1} = \frac{50 \pm 10 \sqrt{34}}{9} \times \frac{81}{140 \pm 10 \sqrt{34}}

= \frac{(5 \pm \sqrt{34}) \times 9}{14 \pm \sqrt{34}} ?

Commented by MJS last updated on 15/Apr/20

= 9 \frac{(5 \pm \sqrt{34}) (14 \mp \sqrt{34})}{(14 \pm \sqrt{34}) (14 \mp \sqrt{34})} = 9 \frac{36 \pm 9 \sqrt{34}}{162} = 2 \pm \frac{\sqrt{34}}{2}

Commented by jagoll last updated on 15/Apr/20

w h a t t h e c o r r e c t a n s w e r ?

i s e e h a s 2 a n s w e r

Commented by MJS last updated on 15/Apr/20

\sin x + \cos x = \frac{4}{5} has 2 solutions in [0; 2 π [

\Rightarrow 2 correct answers

Commented by jagoll last updated on 15/Apr/20

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

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