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 >

cos(2t)=-sin(t)

解答

cos (2 t) = - sin (t)

解答

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

+1

度数

t = 21 0^{\circ} + 36 0^{\circ} n, t = 33 0^{\circ} + 36 0^{\circ} n, t = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

cos (2 t) = - sin (t)

两边减去

- sin (t)

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

使用三角恒等式改写

cos (2 t) + sin (t)

使用倍角公式:

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

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

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

用替代法求解

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

令

: sin (t) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 1

使用法则

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 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

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

将解分隔开

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

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

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

去除括号:

(- a) = - a

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

数字相加/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}

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

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

去除括号:

(- a) = - a

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

数字相减:

- 1 - 3 = - 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

使用分式法则:

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

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

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

u = sin (t)

代回

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

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

sin (t) = - \frac{1}{2} : t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

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

sin (t) = - \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}

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

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

sin (t) = 1 : t = \frac{π}{2} + 2 πn

sin (t) = 1

sin (t) = 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}

t = \frac{π}{2} + 2 πn

t = \frac{π}{2} + 2 πn

合并所有解

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

作图

流行的例子

sin^2(x)-5cos(x)-5=0

sin^{2} (x) - 5 cos (x) - 5 = 0

(1+tan^2(27*θ))/(1-tan^2(27*θ))=sqrt(2)

\frac{1 + tan ^{2} ( 27 \cdot θ )}{1 - tan ^{2} ( 27 \cdot θ )} = 2

tan (x) = - cot (x)

3cos^2(x)-sin^2(x)=0

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

cos (x) = - 0.6