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#