Question

Sin2x-cos2x=√2 sin3x

Answer 1

Ответ:

$-frac{pi }{4} +2pi n, ~ninmathbb {Z} ;frac{pi }{4}+frac{2pi k}{5} ,~kinmathbb {Z}}.$

Объяснение:

$sin2x-cos2x =sqrt{2} sin3x|: sqrt{2} ;\\frac{1}{sqrt{2} } sin2x-frac{1}{sqrt{2} } cos2x =sin3x;\\sin2x* cos frac{pi }{4} - cos2x*sinfrac{pi }{4} =sin3x;\\sin ( 2x- frac{pi }{4} )-sin3x=0;\\2 sin frac{2x-frac{pi }{4}-3x }{2} * cos frac{2x-frac{pi }{4} +3x}{2} =0;$

$2sin( -frac{x}{2} -frac{pi }{8} ) * cos (frac{5x}{2} -frac{pi }{8} )=0 ;\\-2sin( frac{x}{2} +frac{pi }{8} ) * cos (frac{5x}{2} -frac{pi }{8} )=0;\\sin( frac{x}{2} +frac{pi }{8} ) * cos (frac{5x}{2} -frac{pi }{8} )=0;$

$left [ begin{array}{lcl} {{sin( frac{x}{2}+frac{pi }{8} )=0,} \\ {cos( frac{5x}{2}-frac{pi }{8}) =0;}} end{array} right.Leftrightarrow left [ begin{array}{lcl} {{frac{x}{2}+frac{pi }{8} =pi n,~ninmathbb {Z} } \\ {frac{5x}{2} -frac{pi }{8}=frac{pi }{2} +pi k,~kinmathbb {Z}}} end{array} rightLeftrightarrow$

$left [ begin{array}{lcl} {{frac{x}{2} =-frac{pi }{8}+pi n,~ninmathbb {Z} } \\ {frac{5x}{2} =frac{5pi }{8} +pi k,~kinmathbb {Z}}} end{array} right.Leftrightarrow left [ begin{array}{lcl} {{x=-frac{pi }{4} +2pi n,~ninmathbb {Z}} \\ {x=frac{pi }{4} +frac{2pi k}{5} ,~kinmathbb {Z}}} end{array} right.$

Answer 2

Решение приложено...