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人気のある三角関数 >

solvefor x,(d^3+d^2+1)y=cos(3x)

解

解く

x, (d^{3} + d^{2} + 1) y = cos (3 x)

解

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

解答ステップ

(d^{3} + d^{2} + 1) y = cos (3 x)

辺を交換する

cos (3 x) = (d^{3} + d^{2} + 1) y

三角関数の逆数プロパティを適用する

cos (3 x) = (d^{3} + d^{2} + 1) y

以下の一般解

cos (3 x) = (d^{3} + d^{2} + 1) y

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

3 x = arccos ((d^{3} + d^{2} + 1) y) + 2 πn, 3 x = - arccos ((d^{3} + d^{2} + 1) y) + 2 πn

3 x = arccos ((d^{3} + d^{2} + 1) y) + 2 πn, 3 x = - arccos ((d^{3} + d^{2} + 1) y) + 2 πn

解く

3 x = arccos ((d^{3} + d^{2} + 1) y) + 2 πn : x = \frac{arccos ( ( d ^{3} + d ^{2} + 1 ) y )}{3} + \frac{2 πn}{3}

3 x = arccos ((d^{3} + d^{2} + 1) y) + 2 πn

以下で両辺を割る

3

3 x = arccos ((d^{3} + d^{2} + 1) y) + 2 πn

以下で両辺を割る

3

\frac{3 x}{3} = \frac{arccos ( ( d ^{3} + d ^{2} + 1 ) y )}{3} + \frac{2 πn}{3}

簡素化

x = \frac{arccos ( ( d ^{3} + d ^{2} + 1 ) y )}{3} + \frac{2 πn}{3}

x = \frac{arccos ( ( d ^{3} + d ^{2} + 1 ) y )}{3} + \frac{2 πn}{3}

解く

3 x = - arccos ((d^{3} + d^{2} + 1) y) + 2 πn : x = - \frac{arccos ( ( d ^{3} + d ^{2} + 1 ) y )}{3} + \frac{2 πn}{3}

3 x = - arccos ((d^{3} + d^{2} + 1) y) + 2 πn

以下で両辺を割る

3

3 x = - arccos ((d^{3} + d^{2} + 1) y) + 2 πn

以下で両辺を割る

3

\frac{3 x}{3} = - \frac{arccos ( ( d ^{3} + d ^{2} + 1 ) y )}{3} + \frac{2 πn}{3}

簡素化

x = - \frac{arccos ( ( d ^{3} + d ^{2} + 1 ) y )}{3} + \frac{2 πn}{3}

x = - \frac{arccos ( ( d ^{3} + d ^{2} + 1 ) y )}{3} + \frac{2 πn}{3}

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

グラフ

人気の例

arccos(2x-1)= pi/6

arccos (2 x - 1) = \frac{π}{6}

sin(θ)-cos(θ/2)=0

sin (θ) - cos (\frac{θ}{2}) = 0

tan(x)= 195/867

tan (x) = \frac{195}{867}

solvefor x,cos(qx)=sin(rx)

so l v e f or x, cos (q x) = sin (r x)

sin(216)=sin(θ)

sin (21 6^{\circ}) = sin (θ)