Question

sa se calculeze tg x stiind ca sin 2x =-3/5 iar x apartine (3pi/4, pi)​

Answer 1

sin 2x=2tgx/(1+tg²x)

notand tgx=t , avem

-3/5=2t/(1+t²)

10t=-3-3t²

3t²+10t+3=0

t1,2=(-10+-√(100-36))/6= (-10+-√54)/6= (-10+-8)/6

t1=-3

t2=-1/3

x∈(3π/4; π)⇒tgx∈(0;-1) deci convine t2

tgx=-1/3

Answer 2

Răspuns:

tg(x) = -1/3

Explicație pas cu pas:

Datele problemei:

$Sa \: se \: calculeze\:tg(x) \:stiind\: ca: sin(2x) = \frac{-3}{5} , \: x \:apartine (3pi/4; pi)$.

Folosim urmatoarea formula de legatura dintre tg si sin:

$sin(2x) = \frac{2tg(x)}{1+tg^2(x)}$

$\frac{-3}{5} = \frac{2tg(x)}{1+tg^2(x)} \\ \\ -3*(1+tg^2x) = 5 * 2tg(x)\\ \\ \\-3-3tg^2(x) = 10 tg(x)\\ \\ 3tg^2(x) + 10 tg(x) + 3 = 0\\\\ Notez\: \:tg(x) = t \\ \\ Rescriu \: in \: functie\: de\: t:\\ \\ 3t^2 + 10t + 3 = 0\\\\\Delta =10^2 - 4 * 3 * 3 = 100 - 36 = 64\\t_1, t_2 = \frac{-10 \pm \sqrt{64} }{2*3} = \frac{-10 \pm 8}{6} \\\\t_1 = \frac{-10 - 8}{6} = -3\\t_2 = \frac{-10 + 8}{6} = \frac{-2}{6} =\frac{-1}{3} \\\\Cum \: x \: apartine (3pi/4; pi) => tg(x) \: \:apartine ( 0; -1).$

Observam ca $t_2$ se incadreaza in acest interval.

=> tg(x) = -1/3