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受欢迎的三角函数 >

2sin^2(x)=cos^3(x)tan(x)

解答

2 sin^{2} (x) = cos^{3} (x) tan (x)

解答

x = 2 πn, x = π + 2 πn, x = 0.42707 \dots + 2 πn, x = π - 0.42707 \dots + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 24.46980 \dots^{\circ} + 36 0^{\circ} n, x = 155.53019 \dots^{\circ} + 36 0^{\circ} n

求解步骤

2 sin^{2} (x) = cos^{3} (x) tan (x)

两边减去

cos^{3} (x) tan (x)

2 sin^{2} (x) - cos^{3} (x) tan (x) = 0

使用三角恒等式改写

2 sin^{2} (x) - cos^{3} (x) tan (x)

使用基本三角恒等式:

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

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

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

cos^{3} (x) \frac{sin ( x )}{cos ( x )}

分式相乘:

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

= \frac{sin ( x ) cos ^{3} ( x )}{cos ( x )}

约分:

cos (x)

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

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

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

分解

2 sin^{2} (x) - cos^{2} (x) sin (x) : sin (x) (2 sin (x) - cos^{2} (x))

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

使用指数法则:

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

sin^{2} (x) = sin (x) sin (x)

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

因式分解出通项

sin (x)

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

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

分别求解每个部分

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

2 sin (x) - cos^{2} (x) = 0 : x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

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

使用三角恒等式改写

- cos^{2} (x) + 2 sin (x)

使用毕达哥拉斯恒等式:

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

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

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

- (1 - sin^{2} (x)) : - 1 + sin^{2} (x)

- (1 - sin^{2} (x))

打开括号

= - (1) - (- sin^{2} (x))

使用加减运算法则

- (- a) = a, - (a) = - a

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

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

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

用替代法求解

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

令

: sin (x) = u

- 1 + u^{2} + 2 u = 0

- 1 + u^{2} + 2 u = 0 : u = - 1 + 2, u = - 1 - 2

- 1 + u^{2} + 2 u = 0

改写成标准形式

a x^{2} + b x + c = 0

u^{2} + 2 u - 1 = 0

使用求根公式求解

u^{2} + 2 u - 1 = 0

二次方程求根公式:

若

a = 1, b = 2, c = - 1

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 1) = 22

2^{2} - 4 \cdot 1 \cdot (- 1)

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

数字相加:

4 + 4 = 8

= 8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

u_{1, 2} = \frac{- 2 \pm 2 2}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 2 + 2 2}{2 \cdot 1}, u_{2} = \frac{- 2 - 2 2}{2 \cdot 1}

u = \frac{- 2 + 2 2}{2 \cdot 1} : - 1 + 2

\frac{- 2 + 2 2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 2 + 2 2}{2}

分解

- 2 + 22 : 2 (- 1 + 2)

- 2 + 22

改写为

= - 2 \cdot 1 + 22

因式分解出通项

2

= 2 (- 1 + 2)

= \frac{2 ( - 1 + 2 )}{2}

数字相除:

\frac{2}{2} = 1

= - 1 + 2

u = \frac{- 2 - 2 2}{2 \cdot 1} : - 1 - 2

\frac{- 2 - 2 2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 2 - 2 2}{2}

分解

- 2 - 22 : - 2 (1 + 2)

- 2 - 22

改写为

= - 2 \cdot 1 - 22

因式分解出通项

2

= - 2 (1 + 2)

= - \frac{2 ( 1 + 2 )}{2}

数字相除:

\frac{2}{2} = 1

= - (1 + 2)

取负

- (1 + 2) = - 1 - 2

= - 1 - 2

二次方程组的解是:

u = - 1 + 2, u = - 1 - 2

u = sin (x)

代回

sin (x) = - 1 + 2, sin (x) = - 1 - 2

sin (x) = - 1 + 2, sin (x) = - 1 - 2

sin (x) = - 1 + 2 : x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

sin (x) = - 1 + 2

使用反三角函数性质

sin (x) = - 1 + 2

sin (x) = - 1 + 2

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

sin (x) = - 1 - 2 :

无解

sin (x) = - 1 - 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

合并所有解

x = 2 πn, x = π + 2 πn, x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

以小数形式表示解

x = 2 πn, x = π + 2 πn, x = 0.42707 \dots + 2 πn, x = π - 0.42707 \dots + 2 πn

作图

流行的例子

tan(x-10)cot(20-x)=1

tan (x - 1 0^{\circ}) cot (2 0^{\circ} - x) = 1

3csc(2x)-4sin(2x)=0

3 csc (2 x) - 4 sin (2 x) = 0

3tan(B)-4=tan(B)-2

3 tan (B) - 4 = tan (B) - 2

sin(2x)=5cos(2x)

sin (2 x) = 5 cos (2 x)

5 sec (2 x) + 2 = 0