How do you solve $sinx=sqrt2/2$ ?

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Answer 1 · 2018-03-16T09:09:04

$x = (2npi + pi/4)$ or $(2npi + (3pi)/4)$ , $n -> epsilon -> ZZ$

Given $sin x = (sqrt2/2)$

We know 2 = (sqrt2)^2 = sqrt 2 * sqrt 2 #

$:. sin x = sin sqrt 2 / 2 = cancelsqrt 2 / (cancel sqrt 2 * sqrt 2) = 1/sqrt 2$

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From the above table,
$sin (pi/4) = (sin pi - (pi/4)) = sin ((3pi)/4) = 1/sqrt2$

$x = (pi/4)^c$ or $45^@$ , (3pi) / )^c $or$ 135^@#

Generalizing the same,

$x = (2npi + pi/4)$ or $(2npi + (3pi)/4)$ , $n -> epsilon -> ZZ$

How do you solve sinx=sqrt2/2sinx=sqrt2/2?