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3 よって θ = π π (2) 2(sinθ−cos )θ sin cos 6 2 2 θ θ

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(1)

0≦θ <2π のとき,次の方程式を解け。

(1) 3 sinθ+cosθ =0 2 sin 3 cos 1 0

2 2

θ θ

 

⋅ + ⋅ =

 

 

 

sin 0

6 θ π

 + =

 

  …① 0≦θ <2π のとき 13

6 6 6

π ≦θ + π < π より

① ⇔ , 2

6

θ+ π =π π

したがって 5 11

6 , 6

θ = π π

〔別解〕

両辺をcosθで割って, sin tan cos

θ θ

θ = とおくと 3 sinθ+cosθ =0 3 tanθ + =1 0

1 tan

θ = − 3

よって 5 11

6 , 6

θ = π π

(2) 2(sinθ−cos )θ = 6 1 1

2 2 sin cos 6

2 2

θ θ

  

 

⋅  ⋅ + ⋅ − =

  

 

3

sin 4 2

θ π

 − =

 

  …①

0≦θ <2π のとき 7

4 4 4

π θ π π

− ≦ − < より

① ⇔ 2

4 3 , 3

π π

θ− = π

したがって 7 11

12 , 12

θ = π π

87.三角関数を含む方程式④

(1) 5 11 6 , 6

θ = π π (2) 7 11 12 , 12

θ = π π

(3) 5 3

, , ,

4 2 4 2

π π

θ = π π (4) 4

0, , ,

3 3

θ = π π π

(2)

(3) sin 2θ −cos 2θ =1 2 sin 2 1 cos 2 1 1

2 2

θ θ

  

 ⋅ + ⋅ − =

 

  

 

1

sin 2

4 2

θ π

 − =

 

  …①

0≦θ <2π のとき 15

4 2 4 4

π θ π π

− ≦ − < より

① ⇔ 3 9 11

2 , , ,

4 4 4 4 4

π π

θ − = π π π

したがって 5 3

, , ,

4 2 4 2

π π

θ = π π

〔別解〕

2 sin cosθ θ−(2 cos2θ− =1) 1

cos (sinθ θ−cos )θ =0

よって cosθ =0 または sinθ =cosθ

したがって 5 3

, , ,

4 2 4 2

π π

θ = π π

(4) cos2θ + 3 sin cosθ θ =1 1 cos 2 1

3 sin 2 1

2 2

θ θ

+ + ⋅ =

3 1 1

sin 2 cos 2

2 θ+ 2 θ = 2

1

sin 2

6 2

θ π

 + =

 

  …①

0≦θ <2π のとき 25

6 2 6 6

π ≦ θ+ π < π より

① ⇔ 5 13 17

2 , , ,

6 6 6 6 6

π π

θ + = π π π

したがって 4

0, , ,

3 3

θ = π π π

sinθ =cosθは,sin sin 2 θ =  π −θ

  と変形して解いてもよいですし,

単位円周上で,x座標とy座標が等しくなるところ(2つある)を考えてもよいです。

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