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受欢迎的三角函数 >

cos(2x)+sqrt(3)sin(x)=1

解答

cos (2 x) + 3 sin (x) = 1

解答

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

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 6 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n

求解步骤

cos (2 x) + 3 sin (x) = 1

两边减去

1

cos (2 x) + 3 sin (x) - 1 = 0

使用三角恒等式改写

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

使用倍角公式:

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

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

化简

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

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

用替代法求解

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

令

: sin (x) = u

- 2 u^{2} + u 3 = 0

- 2 u^{2} + u 3 = 0 : u = 0, u = \frac{3}{2}

- 2 u^{2} + u 3 = 0

改写成标准形式

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

- 2 u^{2} + 3 u = 0

使用求根公式求解

- 2 u^{2} + 3 u = 0

二次方程求根公式:

若

a = - 2, b = 3, c = 0

u_{1, 2} = \frac{- 3 \pm ( 3 ) ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

u_{1, 2} = \frac{- 3 \pm ( 3 ) ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

(3)^{2} - 4 (- 2) \cdot 0 = 3

(3)^{2} - 4 (- 2) \cdot 0

使用法则

- (- a) = a

= (3)^{2} + 4 \cdot 2 \cdot 0

(3)^{2} = 3

(3)^{2}

使用根式运算法则:

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

= (3^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

约分:

2

= 1

= 3

4 \cdot 2 \cdot 0 = 0

4 \cdot 2 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 3 + 0

数字相加:

3 + 0 = 3

= 3

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

将解分隔开

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

u = \frac{- 3 + 3}{2 ( - 2 )} : 0

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

去除括号:

(- a) = - a

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

同类项相加:

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

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

去除括号:

(- a) = - a

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

同类项相加:

- 3 - 3 = - 23

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 3}{- 4}

使用分式法则:

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

= \frac{2 3}{4}

约分:

2

= \frac{3}{2}

二次方程组的解是:

u = 0, u = \frac{3}{2}

u = sin (x)

代回

sin (x) = 0, sin (x) = \frac{3}{2}

sin (x) = 0, sin (x) = \frac{3}{2}

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

sin (x) = \frac{3}{2}

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

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

合并所有解

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

作图

流行的例子

tan(2θ)-2cos(θ)=0

tan (2 θ) - 2 cos (θ) = 0

tan(θ)= 12/5 ,pi<θ<(3pi)/2

tan (θ) = \frac{12}{5}, π < θ < \frac{3 π}{2}

sqrt(3)tan(3θ)+1=0,0<= θ<= 2pi

3 tan (3 θ) + 1 = 0, 0 \leq θ \leq 2 π

cos(2x)-sqrt(3)sin(x)=1

cos (2 x) - 3 sin (x) = 1

-2sin(x)-sin(2x)=0

- 2 sin (x) - sin (2 x) = 0