Question

Sinx+cosx=√2cos2x find the value of x.

Answer 1

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$\sin( \alpha ) + \cos( \alpha ) = \sqrt{2} \cos(2 \alpha ) \\ \\ \frac{ \sin( \alpha ) + \cos( \alpha ) }{ \sqrt{2} } = \cos(2 \alpha ) \\ \\ \sin( \alpha ) \times \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } \times \cos( \alpha ) = \cos(2 \alpha ) \\ \\ \sin( \alpha ) \sin( \frac{\pi}{4} ) + \cos( \alpha ) \cos( \frac{\pi}{4} ) = \cos(2 \alpha ) \\ \\ \cos( \frac{\pi}{4} - \alpha ) \: \boxed{or \green{ \: \: \: \cos( \alpha - \frac{\pi}{4} ) \: \: \: }}= \cos(2 \alpha ) \\ \\ \frac{\pi}{4} - \alpha = 2 \alpha \: \: \: or \: \: \orange{ \alpha - \frac{\pi}{4} = 2 \alpha } \\ \\ 3 \alpha = \frac{\pi}{4} \: \: \: or \: \: \orange{ \alpha = - \frac{\pi}{4} } \: \: \\ \\ therefore \\ \\ \boxed{ \green{ \: \alpha = - \frac{\pi}{4} \: \: or \: \: \alpha = \frac{\pi}{12} }} \: \: \: \bold \red {ans..............}$

$\boxed{ \: \green{ \: get \: well \: answer \: and \: be \: happy......... \: \: \: \: }}$