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(1)

0

θ < 2 π

のとき,次の不等式を解け。

(1)

3 sin θ + cos θ < − 1

3 1

2 sin cos 1

2 2

θ θ

 

⋅ + ⋅ < −

 

 

 

sin 1

6 2

θ π

 +  < −

 

 

…①

0

θ < 2 π

のとき

13

6 6 6

π ≦ θ + π < π

より

① ⇔

7 11

6 6 6

π θ < + π < π

したがって

5

π θ < < 3 π

(2)

1

sin cos

6 2

θ π θ

 −  + <

 

 

sin cos cos sin cos 1

6 6 2

π π

θ − θ + θ <

3 1 1

sin cos

2 2 2

θ ⋅ + θ ⋅ <

1

sin 6 2

θ π

 +  <

 

 

…①

0

θ < 2 π

のとき

13

6 6 6

π ≦ θ + π < π

より

① ⇔

6 6 4

π ≦ θ + π < π

または

3 13

4 6 6

π θ < + π < π

したがって

0

12 θ < π

または

7

12 π θ < < 2 π

(3)

( 3 sin θ cos θ )( sin θ + 3 cos θ )

3

3 sin

2

θ + 2 sin cos θ θ 3 cos

2

θ 3

− 3(cos

2

θ − sin

2

θ ) + 2 sin cos θ θ ≧ 3

− 3 cos 2 θ + sin 2 θ ≧ 3

sin 2 θ − 3 cos 2 θ ≧ 3

1 3

2 sin 2 cos 2 3

2 2

θ θ

   

 ⋅ + ⋅ −   

   

   

 

90.三角関数を含む方不等式③

(1)

5

π θ < < 3 π

(2)

0

12 θ < π

または

7

12 π θ < < 2 π

(3)

5

3 6

π ≦ ≦ θ π

または

4 11

3 π θ ≦ ≦ 6 π

(2)

3 sin 2

3 2

θ π

 − 

 

 

…①

0

θ < 2 π

のとき

11

3 2 3 3

π θ π π

− ≦ − <

より

① ⇔

4

3 2 3 3

π ≦ θ − π ≦ π

または

7 10

3 2 3 3

π ≦ θ − π ≦ π

したがって

5

3 6

π ≦ ≦ θ π

または

4 11

3 π θ ≦ ≦ 6 π

参照

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