Решите по матану, пожалуйста:

sin(x/4) + cos(x/4) = 1
cos6x - 2sin(3pi/2 + 2x)

и если можно вот это :(

3x49^x - 16x21^x + 21x9^x < 0

12sin(x/8)cos(x/8)+cos²(x/8)-sin²(x/8)-sin²(x/8)-cos²(x/8)=02sin(x/8)cos(x/8)-2sin²(x/8)=02sin(x/8)*(cos(x/8)-sin(x/8))=0sin(x/8)=0⇒x/8=πn.n∈z⇒x=8πn,n∈zcos(x/8)-sin(x/8)=0/cos(x/8)1-tg(x/8)=0⇒tg(x/8)=1⇒x/8=π/4+πk.k∈z⇒x=2π+8πk,k∈z2cos6x+2cos2x=04cos³2x-3cos2x+2cos2x=04cos³2x-cos2x=0cos2x(4cos²2x-1)=0cos2x=0⇒2x=π/2+πn,n∈z⇒x=π/4+πn/2,n∈z4cos²2x-1=04(1+cos4x)/2=11+cos4x=1/2cos4x=-1/24x=+-2π/3+2πk,k∈zx=+-π/6+πk/2,k∈z33*7^2x-16*7^x*3^x+21*3^2x<0/3^2x3(7/3)^2x-16*(7/3)^x+21<0(7/3)^x=a3a²-16a+21<0D=256-252=4√D=2a1=(16-2)/6=7/3a2=(16+2)/6=37/3<a<37/3<(7/3)^x<31<x<log(7/3)3x∈(1;log(7/3)3)