Question

1) 6sin^2 x+2sin^2 2x=5

 

2) sin 3x + sin 5x = 2(cos^2 2x - sin^2 3x)

Answer 1

1) 
$\displaystyle 6sin^2x+2sin^22x=5\\6( \frac{1-cos2x}{2})+2(1-cos^22x)=5\\3-3cos2x+2-2cos^22x-5=0\\-2cos^22x-3cos2x=0\\-cos2x(2cos2x+3)=0$

a) 
$\displaystyle cos2x=0\\2x= \frac{ \pi }{2}+ \pi n; n\in Z\\x= \frac{ \pi }{4}+ \frac{ \pi n}{2};n\in Z$

b) 
$\displaystyle 2cos2x+3=0\\cos2x=-3/2$
нет корней

2) 
$\displaystyle sin3x+sin5x=2(cos^22x-sin^23x)\\2sin \frac{3x+5x}{2}*cos \frac{3x-5x}{2}=2( \frac{1+cos4x}{2}- \frac{1-cos6x}{2})\\2sin4x*cosx=cos4x+cos6x\\2sin4x*cosx=2cos \frac{4x+6x}{2}cos \frac{4x-6x}{2}\\2sin4x*cosx=2cos5x*cosx\\2cosx(sin4x-cos5x)=0$

a) 
$\displaystyle cosx=0\\x= \frac{ \pi }{2}+ \pi n; n\in Z$

b) 
$\displaystyle sin4x-cos5x=0\\cos( \frac{ \pi }{2}-4x)-cos5x=0\\-2sin \frac{ \pi /2-4x+5x}{2}*sin \frac{ \pi /2-4x-5x}{2}=0\\-2sin( \frac{ \pi }{4}+ \frac{x}{2})*sin( \frac{ \pi }{4}- \frac{9x}{2})=0$

$\displaystyle sin( \frac{ \pi }{4}+ \frac{x}{2})=0\\ \frac{ \pi }{4}+ \frac{x}{2}= \pi n; n\in Z\\ \frac{x}{2}=- \frac{ \pi }{4}+ \pi n; n\in Z\\x=- \frac{ \pi }{2}+2 \pi n; n\ in Z$

$\displaystyle sin( \frac{ \pi }{4}- \frac{9x}{2})=0\\- \frac{9x}{2}=- \frac{ \pi }{4}+ \pi n; n\in Z\\x= \frac{ \pi }{18}+ \frac{2 \pi n}{9}; n\in Z$