Question

Sin^10x+cos^10x=29/16cos^4 2x

Answer 1

(1-cos2a)^5/32+(1+cos2x)^5/32=29/16*cos^42a
(1-cos2x)^5+(1+cos2x)^5=58cos^42x
cos2x=t
(1-t)^5+(1+t)^5=58t^4
1-5t+10t²-10t³+5t^4-t^5+1+5t+10t²+10t³+5t^4+t^5=58t^4
58t^4-10t^4-20t²-2=0
24t^4-10t²-1=0
t²=a
24a²-10a-1=0
D=100+96=196
a1=(10-14)/48=-1/12⇒t²=-1/12 нет решения
a2=(10+14)/48=1/2⇒t²=1/2⇒t=-1/√2 U t=1/√2
cos2x=-1/√2 U cos2x=1/√2
2x=+-3π/4+2πk U 2x=+-π/4+2πk
x=+-3π/8+πk U x=+-π/8+πk,k∈z