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

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