Question #899b3

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Answer 1 · 2018-01-03T20:11:33

$x = (5pi)/12$

$sinx + cosx = 1/sqrt2$

Squaring both sides,
$sin^2x + cos^2x + 2sinxcosx = 1/2$

As $sin^2x + cos^2x = 1$
So,
$2sinxcosx = -1/2$

As $sin2x = 2sinxcosx$

$sin2x = -1/2$

$sin2x = sin((5pi)/6)$

$2x = (5pi)/6$

$x = (5pi)/12$

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