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受欢迎的三角函数 >

27(sin(2t)cos(t)-(sin(t))^3)=0

解答

27 (sin (2 t) cos (t) - (sin (t))^{3}) = 0

解答

t = 2 πn, t = π + 2 πn, t = - 0.95531 \dots + πn, t = 0.95531 \dots + πn

+1

度数

t = 0^{\circ} + 36 0^{\circ} n, t = 18 0^{\circ} + 36 0^{\circ} n, t = - 54.73561 \dots^{\circ} + 18 0^{\circ} n, t = 54.73561 \dots^{\circ} + 18 0^{\circ} n

求解步骤

27 (sin (2 t) cos (t) - (sin (t))^{3}) = 0

使用三角恒等式改写

(- (sin (t))^{3} + cos (t) sin (2 t)) \cdot 27

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

= 27 (- (sin (t))^{3} + cos (t) \cdot 2 sin (t) cos (t))

cos (t) \cdot 2 sin (t) cos (t) = 2 cos^{2} (t) sin (t)

cos (t) \cdot 2 sin (t) cos (t)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cos (t) cos (t) = cos^{1 + 1} (t)

= 2 sin (t) cos^{1 + 1} (t)

数字相加:

1 + 1 = 2

= 2 sin (t) cos^{2} (t)

= 27 (- sin^{3} (t) + 2 cos^{2} (t) sin (t))

(- sin^{3} (t) + 2 cos^{2} (t) sin (t)) \cdot 27 = 0

分解

(- sin^{3} (t) + 2 cos^{2} (t) sin (t)) \cdot 27 : 27 sin (t) (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

(- sin^{3} (t) + 2 cos^{2} (t) sin (t)) \cdot 27

分解

- sin^{3} (t) + 2 cos^{2} (t) sin (t) : sin (t) (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

- sin^{3} (t) + 2 cos^{2} (t) sin (t)

使用指数法则:

a^{b + c} = a^{b} a^{c}

sin^{3} (t) = sin (t) sin^{2} (t)

= - sin (t) sin^{2} (t) + 2 sin (t) cos^{2} (t)

因式分解出通项

sin (t)

= sin (t) (- sin^{2} (t) + 2 cos^{2} (t))

分解

2 cos^{2} (t) - sin^{2} (t) : (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

2 cos^{2} (t) - sin^{2} (t)

将

2 cos^{2} (t) - sin^{2} (t)

改写为

(2 cos (t))^{2} - sin^{2} (t)

2 cos^{2} (t) - sin^{2} (t)

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

= (2)^{2} cos^{2} (t) - sin^{2} (t)

使用指数法则:

a^{m} b^{m} = (ab)^{m}

(2)^{2} cos^{2} (t) = (2 cos (t))^{2}

= (2 cos (t))^{2} - sin^{2} (t)

= (2 cos (t))^{2} - sin^{2} (t)

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(2 cos (t))^{2} - sin^{2} (t) = (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

= (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

= sin (t) (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

= sin (t) (2 cos (t) + sin (t)) (2 cos (t) - sin (t)) \cdot 27

27 sin (t) (2 cos (t) + sin (t)) (2 cos (t) - sin (t)) = 0

分别求解每个部分

sin (t) = 0 or 2 cos (t) + sin (t) = 0 or 2 cos (t) - sin (t) = 0

sin (t) = 0 : t = 2 πn, t = π + 2 πn

sin (t) = 0

sin (t) = 0

的通解

sin (x)

周期表（周期为

2 πn

"）：

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

t = 0 + 2 πn, t = π + 2 πn

t = 0 + 2 πn, t = π + 2 πn

解

t = 0 + 2 πn : t = 2 πn

t = 0 + 2 πn

0 + 2 πn = 2 πn

t = 2 πn

t = 2 πn, t = π + 2 πn

2 cos (t) + sin (t) = 0 : t = arctan (- 2) + πn

2 cos (t) + sin (t) = 0

使用三角恒等式改写

2 cos (t) + sin (t) = 0

在两边除以

cos (t), cos (t) \neq = 0

\frac{2 cos ( t ) + sin ( t )}{cos ( t )} = \frac{0}{cos ( t )}

化简

2 + \frac{sin ( t )}{cos ( t )} = 0

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

2 + tan (t) = 0

2 + tan (t) = 0

将

2

到右边

2 + tan (t) = 0

两边减去

2

2 + tan (t) - 2 = 0 - 2

化简

tan (t) = - 2

tan (t) = - 2

使用反三角函数性质

tan (t) = - 2

tan (t) = - 2

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

t = arctan (- 2) + πn

t = arctan (- 2) + πn

2 cos (t) - sin (t) = 0 : t = arctan (2) + πn

2 cos (t) - sin (t) = 0

使用三角恒等式改写

2 cos (t) - sin (t) = 0

在两边除以

cos (t), cos (t) \neq = 0

\frac{2 cos ( t ) - sin ( t )}{cos ( t )} = \frac{0}{cos ( t )}

化简

2 - \frac{sin ( t )}{cos ( t )} = 0

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

2 - tan (t) = 0

2 - tan (t) = 0

将

2

到右边

2 - tan (t) = 0

两边减去

2

2 - tan (t) - 2 = 0 - 2

化简

- tan (t) = - 2

- tan (t) = - 2

两边除以

- 1

- tan (t) = - 2

两边除以

- 1

\frac{- tan ( t )}{- 1} = \frac{- 2}{- 1}

化简

tan (t) = 2

tan (t) = 2

使用反三角函数性质

tan (t) = 2

tan (t) = 2

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

t = arctan (2) + πn

t = arctan (2) + πn

合并所有解

t = 2 πn, t = π + 2 πn, t = arctan (- 2) + πn, t = arctan (2) + πn

以小数形式表示解

t = 2 πn, t = π + 2 πn, t = - 0.95531 \dots + πn, t = 0.95531 \dots + πn

作图

流行的例子

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\frac{9.03}{sin ( 4 0 ^{\circ} )} = \frac{10}{sin ( x )}

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