4cos⁴x-3cos2x-1=0
cos2x=2cos²x-1
4(cos²x)²-3*(2cos²x-1)-1=0
cos²x=t, t∈[ 0 ; 1 ]
4t²-6t+2=0
2t²-3t+1=0
D=1
t₁=1/2, t₂=1
t₁=1/2
cos²x=1/2, cosx=+-√(1/2), cosx=+-√2/2
1. cosx=√2/2, <u>x₁=+-π/4+2πn, n∈Z</u>
2. cosx=-√2/2, x=+-(π-arccos√2/2)+2πn, n∈Z, <u>x₂=+-3π/4+2πn, n∈z</u>
t₂=1
cos²x=1, cosx=+-1
cosx=-1, <u>x₃=π+2πn, n∈Z</u>
cosx=1, <u>x₄=2πn, n∈Z</u>
1) (1.9+8.1)÷ 3×1/4=10×3 1/4=5/6
2)18/5÷(8/3-7/3)+29/5=18/5÷1/3+29/5=54/5+29/5=83/5