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人気のある三角関数 >

cos^2(x)-3/4 =0,0<= x<= 2pi

解

cos^{2} (x) - \frac{3}{4} = 0, 0 \leq x \leq 2 π

解

x = \frac{π}{6}, x = \frac{11 π}{6}, x = \frac{5 π}{6}, x = \frac{7 π}{6}

+1

度

x = 3 0^{\circ}, x = 33 0^{\circ}, x = 15 0^{\circ}, x = 21 0^{\circ}

解答ステップ

cos^{2} (x) - \frac{3}{4} = 0, 0 \leq x \leq 2 π

置換で解く

cos^{2} (x) - \frac{3}{4} = 0

仮定:

cos (x) = u

u^{2} - \frac{3}{4} = 0

u^{2} - \frac{3}{4} = 0 : u = \frac{3}{2}, u = - \frac{3}{2}

u^{2} - \frac{3}{4} = 0

\frac{3}{4}

を右側に移動します

u^{2} - \frac{3}{4} = 0

両辺に

\frac{3}{4}

を足す

u^{2} - \frac{3}{4} + \frac{3}{4} = 0 + \frac{3}{4}

簡素化

u^{2} = \frac{3}{4}

u^{2} = \frac{3}{4}

x^{2} = f (a)

の場合, 解は

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

u = \frac{3}{4}, u = - \frac{3}{4}

\frac{3}{4} = \frac{3}{2}

\frac{3}{4}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2

= \frac{3}{2}

- \frac{3}{4} = - \frac{3}{2}

- \frac{3}{4}

簡素化

\frac{3}{4} : \frac{3}{2}

\frac{3}{4}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2

= \frac{3}{2}

= - \frac{3}{2}

u = \frac{3}{2}, u = - \frac{3}{2}

代用を戻す

u = cos (x)

cos (x) = \frac{3}{2}, cos (x) = - \frac{3}{2}

cos (x) = \frac{3}{2}, cos (x) = - \frac{3}{2}

cos (x) = \frac{3}{2}, 0 \leq x \leq 2 π : x = \frac{π}{6}, x = \frac{11 π}{6}

cos (x) = \frac{3}{2}, 0 \leq x \leq 2 π

以下の一般解

cos (x) = \frac{3}{2}

cos (x)

2 πn

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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

範囲の解答

0 \leq x \leq 2 π

x = \frac{π}{6}, x = \frac{11 π}{6}

cos (x) = - \frac{3}{2}, 0 \leq x \leq 2 π : x = \frac{5 π}{6}, x = \frac{7 π}{6}

cos (x) = - \frac{3}{2}, 0 \leq x \leq 2 π

以下の一般解

cos (x) = - \frac{3}{2}

cos (x)

2 πn

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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

範囲の解答

0 \leq x \leq 2 π

x = \frac{5 π}{6}, x = \frac{7 π}{6}

すべての解を組み合わせる

x = \frac{π}{6}, x = \frac{11 π}{6}, x = \frac{5 π}{6}, x = \frac{7 π}{6}

グラフ

人気の例

sin^2(x)-sin(x)-cos^2(x)=0

sin^{2} (x) - sin (x) - cos^{2} (x) = 0

12cos^2(x)+sin(x)-6=0

12 cos^{2} (x) + sin (x) - 6 = 0

sin (A) = \frac{3}{5}

tan (θ) = 0.2

sin(x)tan(x)=sin(x),0<= x<= 2pi

sin (x) tan (x) = sin (x), 0 \leq x \leq 2 π