Question

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

Answer 1

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

Answer 2

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