Question

Решите уравнение: cos^2*x/2-sin^2*x/2=0 1+cos2x=2cosx : 35 корней из 2*sin(-855 градусов)

Answer 1

Cos²(x/2) -sin²(x/2) =0 ; * * * cos2α =cos²α -sin²α  * *  *
cos(2*x/2) = 0 ;
cosx =0 ; 
x =π/2 +πn , n∈Z.

1+cos2x =2cosx ;
* * *cos2x =cos²x -sin²x = cos²x -(1-cos²x) = 2cos²x -1⇒1+cos2x=2cos²x
2cos²x =2cosx ;
cosx(cosx -1) =0 ;
[ cosx =0 ; cosx -1=0.⇔[ cosx =0 ;cosx =1.⇔[x =π/2 +πn ; x =2πn , n∈Z.

ответ: π/2 +πn ; 2πn , n∈Z.

35√2*sin(-855°) =35√2 *(-sin855°) = - 35√2 *sin(2*360° +135°) = -35√2sin135° =-35√2 *sin(180 -45°) = -35√2*sin45°)  =-35√2 *(1/√2) = -35.