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受欢迎的三角函数 >

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

解答

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

解答

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

+1

度数

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

- 1

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

使用三角恒等式改写

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

使用倍角公式:

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

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

化简

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

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

对同类项分组

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

1 - 1 = 0

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

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

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

用替代法求解

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

令

: cos (x) = u

2 u^{2} - u 2 = 0

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

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 \cdot 2 \cdot 0}{2 \cdot 2}

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

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

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

(- 2)^{2} = 2

(- 2)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

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

= (2)^{2}

使用根式运算法则:

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

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

使用指数法则:

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

= 2^{\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

= 2

4 \cdot 2 \cdot 0 = 0

4 \cdot 2 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 2 - 0

数字相减:

2 - 0 = 2

= 2

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

同类项相加:

2 + 2 = 22

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

数字相乘:

2 \cdot 2 = 4

= \frac{2 2}{4}

约分:

2

= \frac{2}{2}

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

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

使用法则

- (- a) = a

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

同类项相加:

2 - 2 = 0

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

数字相乘:

2 \cdot 2 = 4

= \frac{0}{4}

使用法则

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

= 0

二次方程组的解是:

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

u = cos (x)

代回

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

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

cos (x) = \frac{2}{2} : x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

cos (x) = \frac{2}{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}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

合并所有解

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

作图

流行的例子

2sin(x+pi/2)+3tan(pi-x)=0

2 sin (x + \frac{π}{2}) + 3 tan (π - x) = 0

sin (x) = 0.27

sec(x/2)=cos(x/2)

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

sec^2(x)-2tan^2(x)=0

sec^{2} (x) - 2 tan^{2} (x) = 0

tan(5x)+sqrt(3)=0

tan (5 x) + 3 = 0