How do you solve $sin x = - \frac{1}{\sqrt{2}}$ $sinx=-1/sqrt2$ ?

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Answer 1 · 2018-04-17T00:51:07

Around $- 0.785398163397$ $-0.785398163397$ radians.

We have:

$sin x = - \frac{1}{\sqrt{2}}$ $sinx=-1/sqrt2$

Apply inverse sine on both sides.

$\Rightarrow x = {sin}^{- 1} (- \frac{1}{\sqrt{2}})$ $=>x=sin^-1(-1/sqrt2)$

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

Answer 2 · 2018-04-17T02:22:51

$x = (π + (2 n π + \frac{π}{4})) or (2 n π - (\frac{π}{4})), n \in 0 \to n$ $x =(pi + (2npi + pi/4)) or (2npi - (pi/4)), n in 0 to n$

$sin x = - (\frac{1}{\sqrt{2}})$ $sin x = -(1/sqrt2)$

![ www.nabla.hr )

$From above tables, sin x = - \frac{1}{\sqrt{2}}, x = 225^{\circ} or 315^{\circ}$ $"From above tables," sin x = -1/sqrt2 , x = 225^@ or 315^@$

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

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

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

Generalising,

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

How do you solve sinx=-1/sqrt2sinx=−1√2sinx=-1/sqrt2?