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 >

2sin^2(u)=1+sin(u),0<= t<2pi

解答

2 sin^{2} (u) = 1 + sin (u), 0 \leq t < 2 π

解答

u = \frac{π}{2}, u = \frac{7 π}{6}, u = \frac{11 π}{6}

+1

度数

u = 9 0^{\circ}, u = 21 0^{\circ}, u = 33 0^{\circ}

求解步骤

2 sin^{2} (u) = 1 + sin (u), 0 \leq t < 2 π

用替代法求解

2 sin^{2} (u) = 1 + sin (u)

令

: sin (u) = v

2 v^{2} = 1 + v

2 v^{2} = 1 + v : v = 1, v = - \frac{1}{2}

2 v^{2} = 1 + v

将

v

para o lado esquerdo

2 v^{2} = 1 + v

两边减去

v

2 v^{2} - v = 1 + v - v

化简

2 v^{2} - v = 1

2 v^{2} - v = 1

将

1

para o lado esquerdo

2 v^{2} - v = 1

两边减去

1

2 v^{2} - v - 1 = 1 - 1

化简

2 v^{2} - v - 1 = 0

2 v^{2} - v - 1 = 0

使用求根公式求解

2 v^{2} - v - 1 = 0

二次方程求根公式:

若

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

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

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

(- 1)^{2} - 4 \cdot 2 (- 1) = 3

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

v_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 \cdot 2}

将解分隔开

v_{1} = \frac{- ( - 1 ) + 3}{2 \cdot 2}, v_{2} = \frac{- ( - 1 ) - 3}{2 \cdot 2}

v = \frac{- ( - 1 ) + 3}{2 \cdot 2} : 1

\frac{- ( - 1 ) + 3}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{1 + 3}{2 \cdot 2}

数字相加:

1 + 3 = 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

v = \frac{- ( - 1 ) - 3}{2 \cdot 2} : - \frac{1}{2}

\frac{- ( - 1 ) - 3}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{1 - 3}{2 \cdot 2}

数字相减:

1 - 3 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{4}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{4}

约分:

2

= - \frac{1}{2}

二次方程组的解是:

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

v = sin (u)

代回

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

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

sin (u) = 1, 0 \leq t < 2 π : u = \frac{π}{2}

sin (u) = 1, 0 \leq t < 2 π

sin (u) = 1

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}

u = \frac{π}{2} + 2 πn

u = \frac{π}{2} + 2 πn

在

0 \leq t < 2 π

范围内的解

u = \frac{π}{2}

sin (u) = - \frac{1}{2}, 0 \leq t < 2 π : u = \frac{7 π}{6}, u = \frac{11 π}{6}

sin (u) = - \frac{1}{2}, 0 \leq t < 2 π

sin (u) = - \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}

u = \frac{7 π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn

u = \frac{7 π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn

在

0 \leq t < 2 π

范围内的解

u = \frac{7 π}{6}, u = \frac{11 π}{6}

合并所有解

u = \frac{π}{2}, u = \frac{7 π}{6}, u = \frac{11 π}{6}

作图

流行的例子

tan(3y+60)tan(2y+5)=1

tan (3 y + 60) tan (2 y + 5) = 1

18^2=8^2+20^2-2*8*20*cos(x)

1 8^{2} = 8^{2} + 2 0^{2} - 2 \cdot 8 \cdot 20 \cdot cos (x)

cot (θ) = - \frac{π}{6}

3tan^2(x)-3tan(x)=0

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

2sin^2(x)+2sin(x)+1=0

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