min-y-9-sin-2-x-4-csc-2-x-3-

Question Number 172613 by cortano1 last updated on 29/Jun/22

m i n y = 9 \sin^{2} x + 4 {c s c}^{2} x + 3

Answered by Rasheed.Sindhi last updated on 29/Jun/22

y = {(3 \sin x + 2 \csc x)}^{2} - 12 + 3

y = {(3 \sin x + 2 \csc x)}^{2} - 9

m i n y = {(0)}^{2} - 9 = - 9

Commented by cortano1 last updated on 29/Jun/22

y e s \dots . m i n = 15

Commented by mr W last updated on 29/Jun/22

p l e a s e r e c h e c k s i r :

i t h i n k

3 \sin x + 2 cosec x = 0 i s n o t p o s s i b l e .

Commented by Rasheed.Sindhi last updated on 29/Jun/22

{Y o u}^{'} r e r i g h t s i r!

Answered by mr W last updated on 29/Jun/22

y = 9 \sin^{2} x + 4 {cosec}^{2} x + 3

⩾ 2 \sqrt{9 \times 4} + 3 = 15 = m i n i m u m

Answered by greougoury555 last updated on 29/Jun/22

f (x) = 9 \sin^{2} x + \frac{4}{\sin^{2} x} + 9

l e t \sin^{2} x = t; t ⩾ 0

y = 9 t + \frac{4}{t} + 3

y^{'} = 9 - \frac{4}{t^{2}} = 0 \Rightarrow \frac{(3 t - 2) (3 t + 2)}{t^{2}} = 0

t^{2} = \frac{2}{3} = \sin^{2} x

f {(x)}_{m i n} = 9 . \frac{2}{3} + 4 . \frac{3}{2} + 3 = 6 + 6 + 3 = 15