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人気のある三角関数 >

16cos^2(x)=12

解

16 cos^{2} (x) = 12

解

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

+1

度

x = 3 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 21 0^{\circ} + 36 0^{\circ} n

解答ステップ

16 cos^{2} (x) = 12

置換で解く

16 cos^{2} (x) = 12

仮定:

cos (x) = u

16 u^{2} = 12

16 u^{2} = 12 : u = \frac{3}{2}, u = - \frac{3}{2}

16 u^{2} = 12

以下で両辺を割る

16

16 u^{2} = 12

以下で両辺を割る

16

\frac{16 u ^{2}}{16} = \frac{12}{16}

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

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

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

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

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

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

すべての解を組み合わせる

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

グラフ

人気の例

sin (2 x) = 0.4

sin (x) = - 0.4

solvefor θ,3cos(θ)=1+cos(θ)

so l v e f or θ, 3 cos (θ) = 1 + cos (θ)

tan(a)= 5/12 ,sin(a)<0,sin(a)+cos(a)

tan (a) = \frac{5}{12}, sin (a) < 0, sin (a) + cos (a)

cot(2x)=sqrt(3),0<= x<= 2pi

cot (2 x) = 3, 0 \leq x \leq 2 π