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5sin^2(x)+8cos(x)=8
8-5sin^2(x)-8cos(x)=0
2.5-5sin^2(x)+5.5-8cos(x)=0
5cos^2(x)+3-8cos(x)=0
cos(x)=(8+-sqrt(64-60))/10=(8+-2)/10= 1 или 0,6
Значит:
x = 2 π n, n ∈ <span>Z
</span>x = 2 π k - arccos(3/5), k ∈<span> Z
</span>x = 2 π k + arccos(3/5), k ∈<span> Z
Но sin(x)>0
Тогда:
</span>x = 2 π k + arccos(3/5), k ∈ Z
Ответ вот это: 12х2у-16хух.
Sin^4+cos^2-cos^4=(sin^2+cos^2)(sin^2-cos^2) + cos^2= cos^2+(sin^2-cos^2)= ((1-cos2)/2) + (sin^2-cos^2)