Question

if sin-¹x -cos-¹x =π/6, the x is equal to​

Answer 1

Answer:

${ \sin(x) }^{ - 1} - { \cos(x) }^{ - 1} = \frac{\pi}{6} \\ { \sin(x) }^{ - 1} = \frac{\pi}{6} + { \cos(x) }^{ - 1} \\ { \sin(x) }^{ - 1} = \frac{\pi}{6} + ( \frac{\pi}{2} - { \sin(x) }^{ - 1} ) \\ = \frac{\pi}{6} + \frac{\pi}{2} - { \sin(x) }^{ - 1} \\ 2 { \sin(x) }^{ - 1} = \frac{\pi + 3\pi}{6} \\ 2 { \sin(x) }^{ - 1} = \frac{4\pi}{6} \\ 2 { \sin(x) }^{ - 1} = \frac{2\pi}{3} \\ { \sin(x) }^{ - 1} = \frac{\pi}{3} \\ x = \frac{ \sqrt{3} }{2}$