Решите уравнение 5cos^2 x-корень из 3 sin2x+3sin^2 x=2

мог бы побольше очков дать

ответ:

(1; -3), (2; 4), (1; 2), (-1; 4), (3; 1)

объяснение:

413.

{2х-3у=11

{5х+у=2 |×3

{2х-3у=11

{15х+3у=6

*+*

{17х=17

{у=2-5х

{х=1

{у=2-5×1=-3

(1; -3)

416.

{5х+у=14 |×2

{3х-2у=-2

{10х+2у=28

{3х-2у=-2

*+*

{13х=26

{у=14-5х

{х=2

{у=14-5×2=4

(2; 4)

419.

{х+3у=7

{х+2у=5 |×(-1)

{х+3у=7

{-х-2у=-5

*+*

{у=2

{х=5-2у=5-2×2=1

(1; 2)

422.

{2х+3у=10

{х-2у=-9 |×(-2)

{2х+3у=10

{-2х+4у=18

*+*

{7у=28

{х=-9+2у

{у=4

{х=-9+2×4=-1

(-1; 4)

425.

{х-у=3 |×(-3)

{3х+4у=2

{-3х+3у=-9

{3х+4у=2

*+*

{7у=-7

{х=3+у

{у=-1

{х=3-1=2

(2; -1)

428.

{х-4у=-1 |×(-3)

{3х-у=8

{-3ч+12у=3

{3х-у=8

*+*

{11у=11

{х=4у-1

{у=1

{х=4×1-1=3

(3; 1)

[tex]y=\big (\frac{sin2x}{x+3}\big )^{1-x}=ln\big (\frac{sin2x}{x+3\frac{x}{y} }\big )^{1-x}=(1-x)\cdot ln\frac{sin2x}{x+3}\; \; ,\; \; \; \; \; \; \; (lnu)'=\frac{u'}{u}{y'}{y}=(1-x)'\cdot ln\frac{sin2x}{x+3}+(1-x)\cdot \big (ln\frac{sin2x}{x+3}\big )'{y'}{y}=-ln\frac{sin2x}{x+3}+(1-x)\cdot \frac{x+3}{sin2x}\cdot \frac{2\, cos2x\cdot (x+3)-sin2x}{(x+3)^2}'=y\cdot \big (-ln\frac{sin2x}{x+3}+(1-x)\cdot \frac{2\, cos2x\cdot (x+3)-sin2x}{(x+3)\cdot sin2x}\big )[/tex]

[tex]y'=\big (\frac{sin2x}{x+3}\big )^{1-x}\cdot \big (-ln\frac{sin2x}{x+3}+(1-x)\cdot \frac{2\, cos2x\cdot (x+3)-sin2x}{(x+3)\cdot sin2x}\big '=\big (\frac{sin2x}{x+3}\big )^{1-x}\cdot \big (-ln\frac{sin2x}{x+3}+2\, (1-x)\cdot ctg2x-\frac{(1-x)}{(x+3)}\big )[/tex]