✨ Best Answer ✨
1<√3sinΘ+cosΘ<√3
両辺を2で割ると
1/2<(√3/2)sinΘ+(1/2)cosΘ<√3/2
1/2<sin(Θ+π/6)<√3/2・・・①
0≦Θ≦π/2よりπ/6≦Θ+π/6≦2π/3
よって求めるΘの範囲は
π/6<Θ<π/3
✨ Best Answer ✨
1<√3sinΘ+cosΘ<√3
両辺を2で割ると
1/2<(√3/2)sinΘ+(1/2)cosΘ<√3/2
1/2<sin(Θ+π/6)<√3/2・・・①
0≦Θ≦π/2よりπ/6≦Θ+π/6≦2π/3
よって求めるΘの範囲は
π/6<Θ<π/3
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