Question #899b3

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Answer 1 · 2018-01-03T20:11:33

$x = \frac{5 π}{12}$ $x = (5pi)/12$

$sin x + cos x = \frac{1}{\sqrt{2}}$ $sinx + cosx = 1/sqrt2$

Squaring both sides,
${sin}^{2} x + {cos}^{2} x + 2 sin x cos x = \frac{1}{2}$ $sin^2x + cos^2x + 2sinxcosx = 1/2$

As ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x + cos^2x = 1$
So,
$2 sin x cos x = - \frac{1}{2}$ $2sinxcosx = -1/2$

As $sin 2 x = 2 sin x cos x$ $sin2x = 2sinxcosx$

$sin 2 x = - \frac{1}{2}$ $sin2x = -1/2$

$sin 2 x = sin (\frac{5 π}{6})$ $sin2x = sin((5pi)/6)$

$2 x = \frac{5 π}{6}$ $2x = (5pi)/6$

$x = \frac{5 π}{12}$ $x = (5pi)/12$

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