For x∈ (0 , (π /2)), if cos- 1((7/2) (1 + cos 2 x) + √(s i n2 x - 48 c o s2 x) sin â¡ x) =x-(cos)- 1(k cos x), then the value of k is equal to

Question

For x∈ (0 , (π /2)), if cos- 1((7/2) (1 + cos 2 x) + √(s i n2 x - 48 c o s2 x) sin â¡ x) =x-(cos)- 1(k cos x), then the value of k is equal to (A) 1 (B) 5 (C) 7 (D) 14

Tardigrade Admin · Accepted Answer

y=cos- 1((7/2) (1 + cos 2 x) + √(s i n2 x - 48 c o s2 x) sin â¡ x) =cos- 1((7 cos x) (cos â¡ x) + √1 - 49 (cos)2 x √1 - (cos)2 x) =cos- 1(cos x)-cos- 1(7 cos â¡ x) =x-cos- 1(7 cos x) Hence, the value of k=7

Q. For $x \in (0, \frac{π}{2}),$ if $co s^{- 1} (\frac{7}{2} (1 + cos 2 x) + (s i n^{2} x - 48 co s^{2} x) s in x)$ $= x - (cos)^{- 1} (k cos x),$

then the value of $k$ is equal to

$1$

$5$

$7$

$14$

$y = co s^{- 1} (\frac{7}{2} (1 + cos 2 x) + (s i n^{2} x - 48 co s^{2} x) s in x)$
$= co s^{- 1} ((7 cos x) (cos x) + 1 - 49 (cos)^{2} x 1 - (cos)^{2} x)$
$= co s^{- 1} (cos x) - co s^{- 1} (7 cos x)$
$= x - co s^{- 1} (7 cos x)$
Hence, the value of $k = 7$

Q. For x∈(0,2π​), if cos−1(27​(1+cos2x)+(sin2x−48cos2x)​sin⁡x) =x−(cos)−1(kcosx), then the value of k is equal to

1

5

7

14

y=cos−1(27​(1+cos2x)+(sin2x−48cos2x)​sin⁡x) =cos−1((7cosx)(cos⁡x)+1−49(cos)2x​1−(cos)2x​) =cos−1(cosx)−cos−1(7cos⁡x) =x−cos−1(7cosx) Hence, the value of k=7

Q. For $x \in (0, \frac{π}{2}),$ if $co s^{- 1} (\frac{7}{2} (1 + cos 2 x) + (s i n^{2} x - 48 co s^{2} x) s in x)$ $= x - (cos)^{- 1} (k cos x),$

then the value of $k$ is equal to

$1$

$5$

$7$

$14$

$y = co s^{- 1} (\frac{7}{2} (1 + cos 2 x) + (s i n^{2} x - 48 co s^{2} x) s in x)$
$= co s^{- 1} ((7 cos x) (cos x) + 1 - 49 (cos)^{2} x 1 - (cos)^{2} x)$
$= co s^{- 1} (cos x) - co s^{- 1} (7 cos x)$
$= x - co s^{- 1} (7 cos x)$
Hence, the value of $k = 7$