Question

sin 5x+sinx+2sin(^2)x=1

Answer 1

2sin(3x)*cos(2x) + 2*(1-cos(2x) )/2 =1, 2sin(3x)*cos(2x) + 1- cos(2x)=1, cos(2x)*(2sin(3x) - 1)=0, cos(2x)=0 или 2sin(3x) - 1=0, 2x=n/2+nl, x=n/4+nl/2. sin(3x)=1/2, 3x= (-1)^m*n/6+nm, x= (-1)^m*n/18+nm/3. n - это число пи, l,m,принадлежат Z.

Answer 2

$sin big(5xbig)+sin x+2sin^2x=1\\2sin bigg(dfrac{5x+x}2bigg)cos bigg(dfrac{5x-x}2bigg)-(1-2sin^2x)=0\\2sin big(3xbig)cos big(2xbig)-cosbig(2xbig)=0\\cosbig(2xbig)Big(2sin big(3xbig)-1Big)=0\\1)~cosbig(2xbig)=0\~~~~~2x=dfrac{pi}2+pi n;~~boxed{boldsymbol{x_1=dfrac{pi}4+dfrac{pi n}2;~~nin mathbb Z}}$

$2)~~2sin big(3xbig)-1=0\~~~~~sin big(3xbig)=dfrac12\\~~~~~3x=dfrac{pi}6+2pi k;~~~~boxed{boldsymbol{x_2=dfrac{pi}{18}+dfrac{2pi k}3;~~kin mathbb Z}}\\~~~~~3x=dfrac{5pi}6+2pi m;~~boxed{boldsymbol{x_3=dfrac{5pi}{18}+dfrac{2pi m}3;~~min mathbb Z}}$

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Использованы формулы

$sin alpha +sin beta =2sin bigg(dfrac{alpha +beta }2bigg)cos bigg(dfrac{alpha -beta }2bigg)\\1-2sin^2alpha =cosbig(2alpha big)$