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人気のある三角関数 >

sin^2(2θ)=1,-180<= θ<= 180

解

sin^{2} (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

解

θ = 4 5^{\circ}, θ = - 13 5^{\circ}, θ = 13 5^{\circ}, θ = - 4 5^{\circ}

+1

ラジアン

θ = \frac{π}{4}, θ = - \frac{3 π}{4}, θ = \frac{3 π}{4}, θ = - \frac{π}{4}

解答ステップ

sin^{2} (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

置換で解く

sin^{2} (2 θ) = 1

仮定:

sin (2 θ) = u

u^{2} = 1

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

規則を適用

1 = 1

= 1

- 1 = - 1

- 1

規則を適用

1 = 1

= - 1

u = 1, u = - 1

代用を戻す

u = sin (2 θ)

sin (2 θ) = 1, sin (2 θ) = - 1

sin (2 θ) = 1, sin (2 θ) = - 1

sin (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ} : θ = 4 5^{\circ}, θ = - 13 5^{\circ}

sin (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

以下の一般解

sin (2 θ) = 1

sin (x)

36 0^{\circ} n

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 θ = 9 0^{\circ} + 36 0^{\circ} n

2 θ = 9 0^{\circ} + 36 0^{\circ} n

解く

2 θ = 9 0^{\circ} + 36 0^{\circ} n : θ = 4 5^{\circ} + 18 0^{\circ} n

2 θ = 9 0^{\circ} + 36 0^{\circ} n

以下で両辺を割る

2

2 θ = 9 0^{\circ} + 36 0^{\circ} n

以下で両辺を割る

2

\frac{2 θ}{2} = \frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

簡素化

\frac{2 θ}{2} = \frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

簡素化

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数を割る:

\frac{2}{2} = 1

= θ

簡素化

\frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 4 5^{\circ} + 18 0^{\circ} n

\frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{9 0 ^{\circ}}{2} = 4 5^{\circ}

\frac{9 0 ^{\circ}}{2}

分数の規則を適用する:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{18 0 ^{\circ}}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= 4 5^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

数を割る:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

範囲の解答

- 18 0^{\circ} \leq θ \leq 18 0^{\circ}

θ = 4 5^{\circ}, θ = - 13 5^{\circ}

sin (2 θ) = - 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ} : θ = 13 5^{\circ}, θ = - 4 5^{\circ}

sin (2 θ) = - 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

以下の一般解

sin (2 θ) = - 1

sin (x)

36 0^{\circ} n

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 θ = 27 0^{\circ} + 36 0^{\circ} n

2 θ = 27 0^{\circ} + 36 0^{\circ} n

解く

2 θ = 27 0^{\circ} + 36 0^{\circ} n : θ = 13 5^{\circ} + 18 0^{\circ} n

2 θ = 27 0^{\circ} + 36 0^{\circ} n

以下で両辺を割る

2

2 θ = 27 0^{\circ} + 36 0^{\circ} n

以下で両辺を割る

2

\frac{2 θ}{2} = \frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

簡素化

\frac{2 θ}{2} = \frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

簡素化

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数を割る:

\frac{2}{2} = 1

= θ

簡素化

\frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 13 5^{\circ} + 18 0^{\circ} n

\frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{27 0 ^{\circ}}{2} = 13 5^{\circ}

\frac{27 0 ^{\circ}}{2}

分数の規則を適用する:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{54 0 ^{\circ}}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= 13 5^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

数を割る:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 13 5^{\circ} + 18 0^{\circ} n

θ = 13 5^{\circ} + 18 0^{\circ} n

θ = 13 5^{\circ} + 18 0^{\circ} n

θ = 13 5^{\circ} + 18 0^{\circ} n

θ = 13 5^{\circ} + 18 0^{\circ} n

範囲の解答

- 18 0^{\circ} \leq θ \leq 18 0^{\circ}

θ = 13 5^{\circ}, θ = - 4 5^{\circ}

すべての解を組み合わせる

θ = 4 5^{\circ}, θ = - 13 5^{\circ}, θ = 13 5^{\circ}, θ = - 4 5^{\circ}

グラフ

人気の例

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100 = \frac{50}{sin ( \frac{x}{2} )}

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