Question

Решить уравнение: 4sin^2 (x) + cos (x) - 3.5

Answer 1

$4sin^2(x)+cos(x)-3,5=0\4-4cos^2(x)+cos(x)-3,5=0\0,5-4cos^2(x)+cos(x)=0\cos(x)=t\-4t^2+t+0,5=0\8t^2-2t-1=0\D_1=1+8=9\t_1= frac{1+3}{8}=frac{1}{2} \t_2=-frac{1}{4} \cos(x)=frac{1}{2} \x=+-frac{pi}{3} +2pi k\cos(x)=-frac{1}{4} \x=pi+-arccos(frac{1}{4} )+2pi k$
k∈Z

Answer 2

Спасибо!

Answer 3

4sin^2 (x) + cos (x) - 3.5=0
4-4cos²x+cosx-3,5=0
cosx=a
4a²-a-0,5=0
D=1+8=9
a1=(1-3)/8=-1/4⇒cosx=-1/4⇒x=π-arccos0,25+2πk,k∈z
a2=(1+3)/8=1/2⇒cosx=1/2⇒x=+-π/3+2πk,k∈z

Answer 4

Спасибо!