If 6sin-1(x2 - 6x + 8.5) = π, then x is equal to

Question

If 6sin-1(x2 - 6x + 8.5) = π, then x is equal to (A) 1 (B) 2 (C) 3 (D) 8

Tardigrade Admin · Accepted Answer

We have, 6 sin-1 (x2 - 6x + 8.5) = π ⇒ sin-1 (x2 - 6x + 8.5) = (π/6) ⇒ x2 - 6x + 8.5 = sin (π /6) = (1/2) ⇒ x2 - 6x + 8 = 0 ⇒ (x-4)(x-2) = 0 ⇒ x = 4 or x = 2

Q. If $6 s i n^{- 1} (x^{2} - 6 x + 8.5) = π$ , then $x$ is equal to

$1$

$2$

$3$

$8$

We have, $6 s i n^{- 1} (x^{2} - 6 x + 8.5) = π$
$\Rightarrow s i n^{- 1} (x^{2} - 6 x + 8.5) = \frac{π}{6}$
$\Rightarrow x^{2} - 6 x + 8.5 = s in \frac{π}{6} = \frac{1}{2}$
$\Rightarrow x^{2} - 6 x + 8 = 0$
$\Rightarrow (x - 4) (x - 2) = 0$
$\Rightarrow x = 4$ or $x = 2$

Q. If 6sin−1(x2−6x+8.5)=π, then x is equal to

1

2

3

8

We have, 6sin−1(x2−6x+8.5)=π ⇒sin−1(x2−6x+8.5)=6π​ ⇒x2−6x+8.5=sin6π​=21​ ⇒x2−6x+8=0 ⇒(x−4)(x−2)=0 ⇒x=4 or x=2

Q. If $6 s i n^{- 1} (x^{2} - 6 x + 8.5) = π$ , then $x$ is equal to

$1$

$2$

$3$

$8$

We have, $6 s i n^{- 1} (x^{2} - 6 x + 8.5) = π$
$\Rightarrow s i n^{- 1} (x^{2} - 6 x + 8.5) = \frac{π}{6}$
$\Rightarrow x^{2} - 6 x + 8.5 = s in \frac{π}{6} = \frac{1}{2}$
$\Rightarrow x^{2} - 6 x + 8 = 0$
$\Rightarrow (x - 4) (x - 2) = 0$
$\Rightarrow x = 4$ or $x = 2$