Question #077dd

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Answer 1 · 2018-02-17T13:01:58

See the explaination

#sqrt2cosx = sin2x#

You can use the identity

#sin2x = 2sinxcosx#

You get,

#sqrt2cosx = 2sinxcosx#

#sqrt(2)cosx - 2sinxcosx = 0 #

#cosx ( sqrt(2) - 2sinx ) = 0#

#sqrt2 - 2sinx = 0 => sqrt(2) = 2sinx #

and

#cosx = 0 #

#sqrt2/2 = sinx#

#1/sqrt2 = sinx#

We know , #sin45° = 1/sqrt2#

So, x = 45°

And we know #cos90^circ = 0 #

So #x = 90^circ # also

#=> x = {45^circ , 90^circ } #