If √2 sin 2 x+(3 √2+1) sin x+30 and x2-7 x+10

Question

If √2 sin 2 x+(3 √2+1) sin x+3>0 and x2-7 x+10<0, then x lies in the interval (A) ((-π/4), (3 π/4)) (B) (2, (5 π/4)) (C) (0, (3 π/2)) (D) ((5 π/4), 5)

Tardigrade Admin · Accepted Answer

We have, √2 sin 2 x+(3 √2+1) sin x+3>0 ⇒ (√2 sin x+1)( sin x+3)>0 ⇒ sin x>(-1/√2) x ∈(π, (5 π/4)) ....(i) and x2-7 x+10 < 0 (x-5)(x-2)<0 x ∈(2,5) ....(ii) From Eqs. (i) and (ii), we get x ∈(2, (5 π/4))

Q. If $2 sin^{2} x + (32 + 1) sin x + 3 > 0$ and
$x^{2} - 7 x + 10 < 0$ , then $x$ lies in the interval

$(\frac{- π}{4}, \frac{3 π}{4})$

$(2, \frac{5 π}{4})$

$(0, \frac{3 π}{2})$

$(\frac{5 π}{4}, 5)$

We have,
$2 sin^{2} x + (32 + 1) sin x + 3 > 0$
$\Rightarrow (2 sin x + 1) (sin x + 3) > 0$
$\Rightarrow sin x > \frac{- 1}{2} x \in (π, \frac{5 π}{4}) .... (i)$
and $x^{2} - 7 x + 10 < 0$
$(x - 5) (x - 2) < 0$
$x \in (2, 5) .... (ii)$
From Eqs. (i) and (ii), we get
$x \in (2, \frac{5 π}{4})$

Q. If 2​sin2x+(32​+1)sinx+3>0 and x2−7x+10<0, then x lies in the interval

(4−π​,43π​)

(2,45π​)

(0,23π​)

(45π​,5)