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人気のある三角関数 >

solvefor x,d^2+3d+2y=4cos^2(x)

解

解く

x, d^{2} + 3 d + 2 y = 4 cos^{2} (x)

解

x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

解答ステップ

d^{2} + 3 d + 2 y = 4 cos^{2} (x)

辺を交換する

4 cos^{2} (x) = d^{2} + 3 d + 2 y

置換で解く

4 cos^{2} (x) = d^{2} + 3 d + 2 y

仮定:

cos (x) = u

4 u^{2} = d^{2} + 3 d + 2 y

4 u^{2} = d^{2} + 3 d + 2 y : u = \frac{d ^{2} + 3 d + 2 y}{2}, u = - \frac{d ^{2} + 3 d + 2 y}{2}

4 u^{2} = d^{2} + 3 d + 2 y

以下で両辺を割る

4

4 u^{2} = d^{2} + 3 d + 2 y

以下で両辺を割る

4

\frac{4 u ^{2}}{4} = \frac{d ^{2}}{4} + \frac{3 d}{4} + \frac{2 y}{4}

簡素化

\frac{4 u ^{2}}{4} = \frac{d ^{2}}{4} + \frac{3 d}{4} + \frac{2 y}{4}

簡素化

\frac{4 u ^{2}}{4} : u^{2}

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

数を割る:

\frac{4}{4} = 1

= u^{2}

簡素化

\frac{d ^{2}}{4} + \frac{3 d}{4} + \frac{2 y}{4} : \frac{d ^{2} + 3 d + 2 y}{4}

\frac{d ^{2}}{4} + \frac{3 d}{4} + \frac{2 y}{4}

規則を適用

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

= \frac{d ^{2} + 3 d + 2 y}{4}

u^{2} = \frac{d ^{2} + 3 d + 2 y}{4}

u^{2} = \frac{d ^{2} + 3 d + 2 y}{4}

u^{2} = \frac{d ^{2} + 3 d + 2 y}{4}

x^{2} = f (a)

の場合, 解は

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

u = \frac{d ^{2} + 3 d + 2 y}{4}, u = - \frac{d ^{2} + 3 d + 2 y}{4}

簡素化

\frac{d ^{2} + 3 d + 2 y}{4} : \frac{d ^{2} + 3 d + 2 y}{2}

\frac{d ^{2} + 3 d + 2 y}{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{d ^{2} + 3 d + 2 y}{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{d ^{2} + 3 d + 2 y}{2}

簡素化

- \frac{d ^{2} + 3 d + 2 y}{4} : - \frac{d ^{2} + 3 d + 2 y}{2}

- \frac{d ^{2} + 3 d + 2 y}{4}

簡素化

\frac{d ^{2} + 3 d + 2 y}{4} : \frac{d ^{2} + 3 d + 2 y}{2}

\frac{d ^{2} + 3 d + 2 y}{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{d ^{2} + 3 d + 2 y}{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{d ^{2} + 3 d + 2 y}{2}

= - \frac{2 y + d ^{2} + 3 d}{2}

= - \frac{d ^{2} + 3 d + 2 y}{2}

u = \frac{d ^{2} + 3 d + 2 y}{2}, u = - \frac{d ^{2} + 3 d + 2 y}{2}

代用を戻す

u = cos (x)

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2}, cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2}

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2}, cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2}

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

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2}

三角関数の逆数プロパティを適用する

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2}

以下の一般解

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2}

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

x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

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

cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2}

三角関数の逆数プロパティを適用する

cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2}

以下の一般解

cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2}

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

x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

すべての解を組み合わせる

x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

グラフ

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