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

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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

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

$sin x = -(1/sqrt2)$

![ www.nabla.hr )

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

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

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

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

Generalising,

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

How do you solve sinx=-1/sqrt2sinx=-1/sqrt2?