Решить систему неравенст методом крамера:
3x-5 > 23-4x
7x+3 < 9x-1

1) cos 3x - √2/2 = 0⇒cos3x=√2/2⇒3x=+-π/4+2πn⇒x=+-π/12+2πn/3
2) tg (x/2 + п/4) = -1⇒x/2+π/4=-π/4+πn⇒x/2=-π/2+πn⇒x=-π+2πn
3) sin 2x + cos x = 0⇒2sinxcosx+cosx=0⇒cosx(2sinx+1)=0
cosx=0⇒x=π/2+πn U sinx=-1/2⇒x=(-1)^n+1 *π/6+πn
4) cos 7x + cos x = 0⇒2cos4xcos3x=0
cos4x=0⇒4x=π/2+πn⇒x=π/8+πn/4 U cos3x=0⇒3x=π/2+πn⇒x=π/6+πn/3
5) 2cos²x + 2sinx = 2,5
2(1-sin²x)+ 2sinx = 2,5
2sin²x-2sinx+0,5=0
sinx=a
2a²-2a+0,5=0
4a²-4a+1=0
(2a-1)²=0
a=1/2⇒sinx=1/2⇒x=(-1)^n *π/6+πn
6) 1 + 2sin 2x + 2cos²x = 0
cos²x+sin²x+4sinxcosx+2cos²x=0
sin²x+4sinxcosx+3cos²x=0 /cos²x≠0
tg²x+4tgx+3=0
tgx=a
a²+4a+3=0
a1+a2=-4 U a1*a2=3
a1=-3⇒tgx=-3⇒x=-arctg3+πn
a2=-1⇒tgx=-1⇒x=-π/4+πn

1)cos3x=√2÷2; 3x=arccos√2/2=+-π/4+2πn; n∈z; x=π/12+2πn/3; n∈z   
2)x/2+π/4=-arctg1=-π/4+πn; n∈z; x=π+2πn; n∈z
3)2*sinx*cosx+cosx=0; cosx*(2sinx+1)=0; cosx=0; x=π/2+πn; n∈z
2sinx+1=0; sinx=-0.5; x=-arcsin0,5=-π/6+2πn; n∈z; x=7π/6+2πn; n∈z
4)cos7x+cosx=2cos((7x+x)/2)cos((7x-x)/2)=2cos4x*cos3x=0; cos4x=0; 4x=π/2+πn; n∈z; x=π/8+πn/4; n∈z; cos3x=0; 3x=π/2+πn; n∈z; x=π/6+πn/3; n∈z 
5)2cos²x=1+cos2x=2-2sin²x; 2(1-sin²x+sinx)=2,5; 1-sin²x+sinx=1.25; sinx-sin²x=0.25; sinx-sin²x-0.25=0; -sin²x+sinx+0.25=0; sinx=t; -1≤t≤1; -t²+t+0.25=0; -4t²+4t+1=0; D=32;
(-4+4√2)/-8=(1-√2)/2; (-4-4√2)/-8=(1+√2)/2-посторонний корень; sinx=(1-√2)/2; x=arcsin((1-√2)/2)+2πn; n∈z