一些三角方程1,6sinx+8cosx=52,sin((x/2)-(pi/6))=根号三/2

Question

一些三角方程
1,6sinx + 8cosx = 5
2,sin((x/2)-(pi/6)) = 根号三/2

Answer 1

1.设siny=8/10,则cosy=6/10,y=arcsin(8/10).
∴6sinx + 8cosx =10(6/10sinx+8/10cosx)
=10(cosysinx+sinycosx)
=10sin(x+y)
∴10sin(x+y)=5
∴sin(x+y)=1/2
故x=2kπ+π/6-arcsin(8/10),
或x=2kπ+5π/6-arcsin(8/10),(k=0,±1,±2.).
2.∵sin((x/2)-(π/6))=√3/2
∴(x/2)-(π/6)=2kπ+π/3,
或(x/2)-(π/6)=2kπ+2π/3,(k=0,±1,±2.).
故x=4kπ+2π/3+π/3=(4k+1)π
或x=4kπ+4π/3+π/3=(4k+5/3)π,(k=0,±1,±2.).