<span>cos 3x = √3 / 2
</span>cos 3x = cos п/ 6
3x = -/ + п/ 6 + 2пk , k Є Z
x = -/ + п/ 18 + 2пk/3 , k Є Z
Y=kx+m
{3=k*4+m
{15=k*6+m
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12=2k k=6 m=3-24=-21
ответ y=6x-21
B1=3,b2=√3
q=b2/b1=√3/3
S=b1/(1-q)=3:(1-√3/3)=3:(3-√3)/3=3*3/(3-√3)=3*3/√3(√3-1)=3√3/(√3-1)=
=3√3(√3+1)/2
Sin 4x = 2sin 2x*cos 2x
4sin^2 (2x) - 2sin 2x*cos 2x = 3
4sin^2 (2x) - 2sin 2x*cos 2x = 3sin^2 (2x) + 3cos^2 (2x)
sin^2 (2x) - 2sin 2x*cos 2x - 3cos^2 (2x) = 0
Делим всё на cos^2 (2x), которое точно не равно нулю.
tg^2 (2x) - 2tg 2x - 3 = 0
(tg 2x + 1)(tg 2x - 3) = 0
1) tg 2x = -1, 2x = -pi/4 + pi*k, x = -pi/8 + pi/2*k
2) tg 2x = 3, 2x = arctg 3 + pi*n, x = 1/2*arctg 3 + pi/2*n