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人気のある三角関数 >

3cos^2(x)=1

解

3 cos^{2} (x) = 1

解

x = 0.95531 \dots + 2 πn, x = 2 π - 0.95531 \dots + 2 πn, x = 2.18627 \dots + 2 πn, x = - 2.18627 \dots + 2 πn

+1

度

x = 54.73561 \dots^{\circ} + 36 0^{\circ} n, x = 305.26438 \dots^{\circ} + 36 0^{\circ} n, x = 125.26438 \dots^{\circ} + 36 0^{\circ} n, x = - 125.26438 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

3 cos^{2} (x) = 1

置換で解く

3 cos^{2} (x) = 1

仮定:

cos (x) = u

3 u^{2} = 1

3 u^{2} = 1 : u = \frac{1}{3}, u = - \frac{1}{3}

3 u^{2} = 1

以下で両辺を割る

3

3 u^{2} = 1

以下で両辺を割る

3

\frac{3 u ^{2}}{3} = \frac{1}{3}

簡素化

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3}

x^{2} = f (a)

の場合, 解は

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

u = \frac{1}{3}, u = - \frac{1}{3}

代用を戻す

u = cos (x)

cos (x) = \frac{1}{3}, cos (x) = - \frac{1}{3}

cos (x) = \frac{1}{3}, cos (x) = - \frac{1}{3}

cos (x) = \frac{1}{3} : x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (x) = \frac{1}{3}

三角関数の逆数プロパティを適用する

cos (x) = \frac{1}{3}

以下の一般解

cos (x) = \frac{1}{3}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (x) = - \frac{1}{3} : x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

cos (x) = - \frac{1}{3}

三角関数の逆数プロパティを適用する

cos (x) = - \frac{1}{3}

以下の一般解

cos (x) = - \frac{1}{3}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

すべての解を組み合わせる

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn, x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

10進法形式で解を証明する

x = 0.95531 \dots + 2 πn, x = 2 π - 0.95531 \dots + 2 πn, x = 2.18627 \dots + 2 πn, x = - 2.18627 \dots + 2 πn

グラフ

人気の例

cos(θ)=-7/8 ,csc(θ)

cos (θ) = - \frac{7}{8}, csc (θ)

1+2sin(x)-cos^2(x)=0

1 + 2 sin (x) - cos^{2} (x) = 0

tan(θ)=-3,sin(θ)<0

tan (θ) = - 3, sin (θ) < 0

solvefor x,sec(x)+sqrt(2)=0

so l v e f or x, sec (x) + 2 = 0

cos (2 x) = 0.92