How do you solve #sinx=-1/sqrt2#?

Question

28947 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-17T00:51:07

Around #-0.785398163397# radians.

We have:

#sinx=-1/sqrt2#

Apply inverse sine on both sides.

#=>x=sin^-1(-1/sqrt2)#

We find this to be around #-0.785398163397# radians.

Answer 2 · 2018-04-17T02:22:51

#x =(pi + (2npi + pi/4)) or (2npi - (pi/4)), n in 0 to n#

#sin x = -(1/sqrt2)#

#"From above tables," sin x = -1/sqrt2 , x = 225^@ or 315^@#

#color(indigo)( " as sin is negative in third and fourth quadrants"#

#:. x = sin ^-1 (-1/sqrt2) = (180 + 45)^@ or (360-45)^@#

#x = (pi + (pi/4))^c or (2pi - (pi/4))^c#

Generalising,

#x =(pi + (2npi + pi/4)) or (2npi - (pi/4)), n in 0 to n#