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受欢迎的三角函数 >

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

解答

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

解答

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

cos (2 x)

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

使用三角恒等式改写

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

使用倍角公式:

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

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

化简

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

- 1 - (2 cos^{2} (x) - 1) + 3 cos (x)

- (2 cos^{2} (x) - 1) : - 2 cos^{2} (x) + 1

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

打开括号

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

使用加减运算法则

- (- a) = a, - (a) = - a

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

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

化简

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

- 1 - 2 cos^{2} (x) + 1 + 3 cos (x)

对同类项分组

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

- 1 + 1 = 0

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

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

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

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

用替代法求解

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

令

: cos (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 = cos (x)

代回

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

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

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

cos (x) = \frac{3}{2} : x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

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

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

作图

流行的例子

7cos^2(θ)-9=-9sin(θ)

7 cos^{2} (θ) - 9 = - 9 sin (θ)

solvefor y,arcsin(y)=x

so l v e f or y, arcsin (y) = x

-4cos(2θ)-3cos(θ)-4=-2cos(θ)-7

- 4 cos (2 θ) - 3 cos (θ) - 4 = - 2 cos (θ) - 7

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

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

2cos^2(x)-9cos(x)+4=0

2 cos^{2} (x) - 9 cos (x) + 4 = 0