Question

Решить уравнения 1.2sin(п/3-x/4)=√3 2.cos(п/2-x/2)-3 cos(п-x/2)=0 3.sin²x+√3 sin x·cos x=0 4.2 sin²x+5sin x·cos x-7 cos²x=0 нужноо. заранее

Answer 1

$1. 2sin( \frac{ \pi }{3} - \frac{x}{4} )= \sqrt{3} \\ sin( \frac{ \pi }{3}- \frac{x}{4} )= \frac{ \sqrt{3} }{2} \\ \left \{ {{\frac{ \pi }{3} - \frac{x}{4} = \frac{ \pi }{3}+2 \pi n } \atop {\frac{ \pi }{3} - \frac{x}{4} = \frac{2 \pi }{3}+2 \pi n }} \right. \\ \left \{ {{- \frac{x}{4} = 2 \pi n } \atop {- \frac{x}{4} = \frac{ \pi }{3}+2 \pi n }} \right. \\ \left \{ {{ \frac{x}{4} =-2 \pi n} \atop { \frac{x}{4} =- \frac{ \pi }{3} -2 \pi n}} \right.$
$\left \{ {{x=-8 \pi n} \atop {x=- \frac{4 \pi }{3} -8 \pi n}} \right. \\ OTBET:-8 \pi n;- \frac{4 \pi }{3}-8 \pi n$ n∈Z

$2.cos( \frac{ \pi }{2}- \frac{x}{2} )-3cos( \pi - \frac{x}{2} )=0 \\ sin \frac{x}{2} +3cos \frac{x}{2} =0|:cos \frac{x}{2} \\ tg \frac{x}{2} +3=0 \\ tg \frac{x}{2} =-3 \\ \frac{x}{2} =-arctg3+ \pi n \\ x=-2arctg3+2 \pi n=2( \pi n-arctg3) \\ OTBET:2 \pi n-arctg3$ n∈Z

$3.sin^2x+ \sqrt{3}sinxcosx=0|:sin^2x \\ 1+ \sqrt{3} ctgx=0 \\ ctgx=- \frac{1}{ \sqrt{3} } \\ x= \frac{2 \pi }{3} + \pi n \\ OTBET: \frac{2 \pi }{3} + \pi n$ n∈Z

$4. 2 sin^2x+5sin xcos x-7 cos^2x=0|:cos^2x \\ 2tg^2x+5tgx-7=0 \\ t=tgx \\ 2t^2+5t-7=0 \\ \left \{ {{t_1=1} \atop {t_2=-3,5}} \right. \\ \left \{ {{tgx=1} \atop {tgx=-3,5}} \right. \\ \left \{ {{x= \frac{ \pi }{4}+ \pi n } \atop {x=-arctg3,5+ \pi n}} \right. \\ OTBET: \frac{ \pi }{4} + \pi n;-arctg3,5+ \pi n$ n∈Z