MQ34 --- Trigonometry
回答 (1)
2012-07-31 7:47 pm
✔ 最佳答案
Solve √3=Cos2x+√3Sin2x which 0<=x<=π/2Sol
√(1^2+(√3)^2)=2
√3=Cos2x+√3Sin2x
√3/2=(1/2)*Cos2x+(√3/2)*Sin2x
√3/2=Sin(π/6)*Cos2x+Cos(π/6)*Sin2x
√3/2=Sin(2x+π/6)
Sin(π/3)=Sin(2x+π/6)
0<=x<=π/2
0<=2x<=π
π/6<=2x+π/6<=7π/6
2x+π/6=π/3 or 2x+π/6=2π/3
2x=π/6 or 2x=π/2
x=π/12 or x=π/4
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https://hk.answers.yahoo.com/question/index?qid=20120731000051KK00178