zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

4sin^4(x)+12cos^2(x)-7=0

解答

4 sin^{4} (x) + 12 cos^{2} (x) - 7 = 0

解答

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

+1

度数

x = 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

求解步骤

4 sin^{4} (x) + 12 cos^{2} (x) - 7 = 0

使用三角恒等式改写

- 7 + 12 cos^{2} (x) + 4 sin^{4} (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)

= - 7 + 12 (1 - sin^{2} (x)) + 4 sin^{4} (x)

化简

- 7 + 12 (1 - sin^{2} (x)) + 4 sin^{4} (x) : 4 sin^{4} (x) - 12 sin^{2} (x) + 5

- 7 + 12 (1 - sin^{2} (x)) + 4 sin^{4} (x)

乘开

12 (1 - sin^{2} (x)) : 12 - 12 sin^{2} (x)

12 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 12, b = 1, c = sin^{2} (x)

= 12 \cdot 1 - 12 sin^{2} (x)

数字相乘:

12 \cdot 1 = 12

= 12 - 12 sin^{2} (x)

= - 7 + 12 - 12 sin^{2} (x) + 4 sin^{4} (x)

数字相加/相减:

- 7 + 12 = 5

= 4 sin^{4} (x) - 12 sin^{2} (x) + 5

= 4 sin^{4} (x) - 12 sin^{2} (x) + 5

5 - 12 sin^{2} (x) + 4 sin^{4} (x) = 0

用替代法求解

5 - 12 sin^{2} (x) + 4 sin^{4} (x) = 0

令

: sin (x) = u

5 - 12 u^{2} + 4 u^{4} = 0

5 - 12 u^{2} + 4 u^{4} = 0 : u = \frac{5}{2}, u = - \frac{5}{2}, u = \frac{1}{2}, u = - \frac{1}{2}

5 - 12 u^{2} + 4 u^{4} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

4 u^{4} - 12 u^{2} + 5 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

4 v^{2} - 12 v + 5 = 0

解

4 v^{2} - 12 v + 5 = 0 : v = \frac{5}{2}, v = \frac{1}{2}

4 v^{2} - 12 v + 5 = 0

使用求根公式求解

4 v^{2} - 12 v + 5 = 0

二次方程求根公式:

若

a = 4, b = - 12, c = 5

v_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 4 \cdot 5}{2 \cdot 4}

v_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 4 \cdot 5}{2 \cdot 4}

(- 12)^{2} - 4 \cdot 4 \cdot 5 = 8

(- 12)^{2} - 4 \cdot 4 \cdot 5

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 12)^{2} = 1 2^{2}

= 1 2^{2} - 4 \cdot 4 \cdot 5

数字相乘:

4 \cdot 4 \cdot 5 = 80

= 1 2^{2} - 80

1 2^{2} = 144

= 144 - 80

数字相减:

144 - 80 = 64

= 64

因式分解数字:

64 = 8^{2}

= 8^{2}

使用根式运算法则:

n a^{n} = a

8^{2} = 8

= 8

v_{1, 2} = \frac{- ( - 12 ) \pm 8}{2 \cdot 4}

将解分隔开

v_{1} = \frac{- ( - 12 ) + 8}{2 \cdot 4}, v_{2} = \frac{- ( - 12 ) - 8}{2 \cdot 4}

v = \frac{- ( - 12 ) + 8}{2 \cdot 4} : \frac{5}{2}

\frac{- ( - 12 ) + 8}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{12 + 8}{2 \cdot 4}

数字相加:

12 + 8 = 20

= \frac{20}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{20}{8}

约分:

4

= \frac{5}{2}

v = \frac{- ( - 12 ) - 8}{2 \cdot 4} : \frac{1}{2}

\frac{- ( - 12 ) - 8}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{12 - 8}{2 \cdot 4}

数字相减:

12 - 8 = 4

= \frac{4}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{4}{8}

约分:

4

= \frac{1}{2}

二次方程组的解是:

v = \frac{5}{2}, v = \frac{1}{2}

v = \frac{5}{2}, v = \frac{1}{2}

代回

v = u^{2}

，求解

u

解

u^{2} = \frac{5}{2} : u = \frac{5}{2}, u = - \frac{5}{2}

u^{2} = \frac{5}{2}

对于

x^{2} = f (a)

解为

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

u = \frac{5}{2}, u = - \frac{5}{2}

解

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

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

对于

x^{2} = f (a)

解为

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

u = \frac{1}{2}, u = - \frac{1}{2}

解为

u = \frac{5}{2}, u = - \frac{5}{2}, u = \frac{1}{2}, u = - \frac{1}{2}

u = sin (x)

代回

sin (x) = \frac{5}{2}, sin (x) = - \frac{5}{2}, sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{5}{2}, sin (x) = - \frac{5}{2}, sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{5}{2} :

无解

sin (x) = \frac{5}{2}

- 1 \leq sin (x) \leq 1

无解

sin (x) = - \frac{5}{2} :

无解

sin (x) = - \frac{5}{2}

- 1 \leq sin (x) \leq 1

无解

sin (x) = \frac{1}{2} : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

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 = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

sin (x) = - \frac{1}{2}

sin (x) = - \frac{1}{2}

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 = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

合并所有解

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

作图

流行的例子

5 sin (a) = 4

2cos^4(x)+8sin^2(x)=5

2 cos^{4} (x) + 8 sin^{2} (x) = 5

cos^4(a)+cos^2(a)=1

cos^{4} (a) + cos^{2} (a) = 1

1-cos(x)=2+cos(x)

1 - cos (x) = 2 + cos (x)

(1+cos(4x))sin(4x)=2cos^2(2x)

(1 + cos (4 x)) sin (4 x) = 2 cos^{2} (2 x)