Question

sin x/2 = -1
tg(2x-п/4) = 1
2 sin^2 x=1+cosx
sin^2 x-5sinx*cosx+4cos^2x=0
cos2x+cosx=0
Cпасибо большое заранее.

Answer 1

k∈Z (k - любое целое число)
$sin \frac{x}{2} =-1 \\ \frac{x}{2}= \frac{3 \pi }{2} \\ x=3 \pi +4 \pi k \\ \\ tg(2x- \frac{ \pi}{4} )=1 \\ 2x- \frac{ \pi }{4}= \frac{ \pi }{4} \\ x= \frac{ \pi }{4} + \frac{ \pi k}{2} \\ \\ 2sin^2x=1+cosx \\ 2\times \frac{1-cos2x}{2} =1+cosx \\ 1-cos2x=1+cosx \\ cos2x=-cosx \\ 2cos^2x+cosx-1=0 \\ cosx=t \\ 2t^2+t-1=0, t_1=-1, t_2=0,5 \\ x= \pi +2 \pi k \\ x= \frac{ \pi }{3} +2 \pi k \\ x= \frac{5 \pi }{3}+2 \pi k$
$sin^2x-5sinxcosx+4cos^2x=0 | : (cos^2x)\\tg^2x-5tgx+4=0 \\ tgx=t \\ t^2-5t+4=0, t_1=1, t_2=4 \\ x= \frac{ \pi }{4}+ \pi k \\ x=arctg 4 + \pi k \\ \\ cos2x+cosx=0 \\ 2cos^2x-1+cosx=0 \\ cosx=t \\ 2t^2+t-1=0, t_1=-1, t_2=0,5 \\ x= \pi +2 \pi k \\ x= \frac{ \pi }{3} +2 \pi k \\ x= \frac{5 \pi }{3}+2 \pi k$