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 >

5sin(x)-3cos(2x)+3=0

解答

5 sin (x) - 3 cos (2 x) + 3 = 0

解答

x = 2 πn, x = π + 2 πn, x = - 0.98511 \dots + 2 πn, x = π + 0.98511 \dots + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = - 56.44269 \dots^{\circ} + 36 0^{\circ} n, x = 236.44269 \dots^{\circ} + 36 0^{\circ} n

求解步骤

5 sin (x) - 3 cos (2 x) + 3 = 0

使用三角恒等式改写

3 - 3 cos (2 x) + 5 sin (x)

使用倍角公式:

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

= 3 - 3 (1 - 2 sin^{2} (x)) + 5 sin (x)

化简

3 - 3 (1 - 2 sin^{2} (x)) + 5 sin (x) : 6 sin^{2} (x) + 5 sin (x)

3 - 3 (1 - 2 sin^{2} (x)) + 5 sin (x)

乘开

- 3 (1 - 2 sin^{2} (x)) : - 3 + 6 sin^{2} (x)

- 3 (1 - 2 sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 3, b = 1, c = 2 sin^{2} (x)

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

使用加减运算法则

- (- a) = a

= - 3 \cdot 1 + 3 \cdot 2 sin^{2} (x)

化简

- 3 \cdot 1 + 3 \cdot 2 sin^{2} (x) : - 3 + 6 sin^{2} (x)

- 3 \cdot 1 + 3 \cdot 2 sin^{2} (x)

数字相乘:

3 \cdot 1 = 3

= - 3 + 3 \cdot 2 sin^{2} (x)

数字相乘:

3 \cdot 2 = 6

= - 3 + 6 sin^{2} (x)

= - 3 + 6 sin^{2} (x)

= 3 - 3 + 6 sin^{2} (x) + 5 sin (x)

3 - 3 = 0

= 6 sin^{2} (x) + 5 sin (x)

= 6 sin^{2} (x) + 5 sin (x)

5 sin (x) + 6 sin^{2} (x) = 0

用替代法求解

5 sin (x) + 6 sin^{2} (x) = 0

令

: sin (x) = u

5 u + 6 u^{2} = 0

5 u + 6 u^{2} = 0 : u = 0, u = - \frac{5}{6}

5 u + 6 u^{2} = 0

改写成标准形式

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

6 u^{2} + 5 u = 0

使用求根公式求解

6 u^{2} + 5 u = 0

二次方程求根公式:

若

a = 6, b = 5, c = 0

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 6 \cdot 0}{2 \cdot 6}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 6 \cdot 0}{2 \cdot 6}

5^{2} - 4 \cdot 6 \cdot 0 = 5

5^{2} - 4 \cdot 6 \cdot 0

使用法则

0 \cdot a = 0

= 5^{2} - 0

5^{2} - 0 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 5

u_{1, 2} = \frac{- 5 \pm 5}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- 5 + 5}{2 \cdot 6}, u_{2} = \frac{- 5 - 5}{2 \cdot 6}

u = \frac{- 5 + 5}{2 \cdot 6} : 0

\frac{- 5 + 5}{2 \cdot 6}

数字相加/相减:

- 5 + 5 = 0

= \frac{0}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{0}{12}

使用法则

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

= 0

u = \frac{- 5 - 5}{2 \cdot 6} : - \frac{5}{6}

\frac{- 5 - 5}{2 \cdot 6}

数字相减:

- 5 - 5 = - 10

= \frac{- 10}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 10}{12}

使用分式法则:

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

= - \frac{10}{12}

约分:

2

= - \frac{5}{6}

二次方程组的解是:

u = 0, u = - \frac{5}{6}

u = sin (x)

代回

sin (x) = 0, sin (x) = - \frac{5}{6}

sin (x) = 0, sin (x) = - \frac{5}{6}

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) = - \frac{5}{6} : x = arcsin (- \frac{5}{6}) + 2 πn, x = π + arcsin (\frac{5}{6}) + 2 πn

sin (x) = - \frac{5}{6}

使用反三角函数性质

sin (x) = - \frac{5}{6}

sin (x) = - \frac{5}{6}

的通解

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

x = arcsin (- \frac{5}{6}) + 2 πn, x = π + arcsin (\frac{5}{6}) + 2 πn

x = arcsin (- \frac{5}{6}) + 2 πn, x = π + arcsin (\frac{5}{6}) + 2 πn

合并所有解

x = 2 πn, x = π + 2 πn, x = arcsin (- \frac{5}{6}) + 2 πn, x = π + arcsin (\frac{5}{6}) + 2 πn

以小数形式表示解

x = 2 πn, x = π + 2 πn, x = - 0.98511 \dots + 2 πn, x = π + 0.98511 \dots + 2 πn

作图

流行的例子

(tan^2(x))/(sec(x)+1)=tan(x)

\frac{tan ^{2} ( x )}{sec ( x ) + 1} = tan (x)

2cos^2(x)+cos(x)=1,0<= x<2pi

2 cos^{2} (x) + cos (x) = 1, 0 \leq x < 2 π

sin(β)=-0,8(θ\in βvc)s=sec(β)-tan(β)

sin (β) = - 0, 8 (θ \in β v c) s = sec (β) - tan (β)

cos((2x-pi)/(17))=0

cos (\frac{2 x - π}{17}) = 0

tan(X)cot(X)-tan(X)+2cot(X)=0

tan (X) cot (X) - tan (X) + 2 cot (X) = 0