Известный румынский математик прошлого века - Б. Угуртов очень просто через логарифмы нашёл решение этой задачи.
*cos2beta = 2cos2beta = psinbeta - 4 = 0
Подставляем логарифм E=log(2) со значением переменной - 2
Получаем 8cos2beta = log(2) 2 cos 2 beta + 9 sin beta (2) = 121.
Со вторым тоже самое.
<span>1)sin(x-pi/4)=0</span>
t=<span>x-pi/4</span>
<span>sint=0</span>
<span>1.t=0+2pi*k</span>
x-pi/4=0
x=pi/4
2.t=pi+2pi*k
pi=x-pi/4
x=5pi/4+2pi*k
3)<span>sinx(3x+pi/3)=1</span>
<span>t=<span>3x+pi/3</span></span>
<span><span>sint=1</span></span>
<span><span>t=pi/2</span></span>
<span><span>3x+pi/3=pi/2</span></span>
<span><span>x=pi/18</span></span>
<span><span>5)<span>tg(x-pi/6)=-√3</span></span></span>
<span><span><span>t=x-pi/6</span></span></span>
<span><span><span>tgt=-√3</span></span></span>
<span><span><span>t=-pi/3+2pi*k . k=Z</span></span></span>
<span><span><span>t=2pi/3+2pi*k . k=Z</span></span></span>
<span>a=(2;3) b=(1;8) c=(4;2)</span>
<span>1) a+b=(2+1;3+8)=(3;11)</span>
<span> Ia+bI=sqrt(3^2+11^2)=sqrt130</span>
<span>2) 2a-b=(2*2-1;2*3-8)=(3;-2)</span>
<span> I2a-bI=sqrt(3^2+(-2)^2)=sqrt13</span>
<span>3) a+2b-3c=(2+2-12;3+16-6)=(-8;13)</span>
<span> Ia+2b-3cI=sqrt(64+169)=sqrt233</span>
<span>4) 4a=(8;12)</span>
<span> I4aI=sqrt(64+144)=sqrt208</span>
<span>5) 3a-8b=(-2;-55)</span>
<span> I3a-8bI=sqrt(4+3025)=sqrt3029</span>
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!