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(x)+cos(2x)=1

解答

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

解答

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

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

1

2 sin (x) + cos (2 x) - 1 = 0

使用三角恒等式改写

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

使用倍角公式:

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

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

化简

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

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

用替代法求解

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

令

: sin (x) = u

2 u - 2 u^{2} = 0

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

2 u - 2 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 2, b = 2, c = 0

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

2^{2} - 4 (- 2) \cdot 0 = 2

2^{2} - 4 (- 2) \cdot 0

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 2 \cdot 0

使用法则

0 \cdot a = 0

= 2^{2} + 0

2^{2} + 0 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 2

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

将解分隔开

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

u = \frac{- 2 + 2}{2 ( - 2 )} : 0

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

去除括号:

(- a) = - a

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

数字相加/相减:

- 2 + 2 = 0

= \frac{0}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{0}{- 4}

使用分式法则:

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

= - \frac{0}{4}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

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

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

去除括号:

(- a) = - a

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

数字相减:

- 2 - 2 = - 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 = 0, u = 1

u = sin (x)

代回

sin (x) = 0, sin (x) = 1

sin (x) = 0, sin (x) = 1

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

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

sin (x) = 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}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

合并所有解

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

作图

流行的例子

2 sin (2 x) = 0

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

sec (θ) - 2 = 0

4sin^2(x)+4cos(x)-5=0

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

2tan(θ)=sin(θ)

2 tan (θ) = sin (θ)