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受欢迎的三角函数 >

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

解答

sin (x) - 2 cos (2 x) = - \frac{1}{2}

解答

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = - 0.84806 \dots + 2 πn, x = π + 0.84806 \dots + 2 πn

+1

度数

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = - 48.59037 \dots^{\circ} + 36 0^{\circ} n, x = 228.59037 \dots^{\circ} + 36 0^{\circ} n

求解步骤

sin (x) - 2 cos (2 x) = - \frac{1}{2}

两边减去

- \frac{1}{2}

sin (x) - 2 cos (2 x) + \frac{1}{2} = 0

化简

sin (x) - 2 cos (2 x) + \frac{1}{2} : \frac{2 sin ( x ) - 4 cos ( 2 x ) + 1}{2}

sin (x) - 2 cos (2 x) + \frac{1}{2}

将项转换为分式:

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

= \frac{sin ( x ) \cdot 2}{2} - \frac{2 cos ( 2 x ) \cdot 2}{2} + \frac{1}{2}

因为分母相等，所以合并分式:

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

= \frac{sin ( x ) \cdot 2 - 2 cos ( 2 x ) \cdot 2 + 1}{2}

数字相乘:

2 \cdot 2 = 4

= \frac{2 sin ( x ) - 4 cos ( 2 x ) + 1}{2}

\frac{2 sin ( x ) - 4 cos ( 2 x ) + 1}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 sin (x) - 4 cos (2 x) + 1 = 0

使用三角恒等式改写

1 + 2 sin (x) - 4 cos (2 x)

使用倍角公式:

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

= 1 + 2 sin (x) - 4 (1 - 2 sin^{2} (x))

化简

1 + 2 sin (x) - 4 (1 - 2 sin^{2} (x)) : 8 sin^{2} (x) + 2 sin (x) - 3

1 + 2 sin (x) - 4 (1 - 2 sin^{2} (x))

乘开

- 4 (1 - 2 sin^{2} (x)) : - 4 + 8 sin^{2} (x)

- 4 (1 - 2 sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 4, b = 1, c = 2 sin^{2} (x)

= - 4 \cdot 1 - (- 4) \cdot 2 sin^{2} (x)

使用加减运算法则

- (- a) = a

= - 4 \cdot 1 + 4 \cdot 2 sin^{2} (x)

化简

- 4 \cdot 1 + 4 \cdot 2 sin^{2} (x) : - 4 + 8 sin^{2} (x)

- 4 \cdot 1 + 4 \cdot 2 sin^{2} (x)

数字相乘:

4 \cdot 1 = 4

= - 4 + 4 \cdot 2 sin^{2} (x)

数字相乘:

4 \cdot 2 = 8

= - 4 + 8 sin^{2} (x)

= - 4 + 8 sin^{2} (x)

= 1 + 2 sin (x) - 4 + 8 sin^{2} (x)

化简

1 + 2 sin (x) - 4 + 8 sin^{2} (x) : 8 sin^{2} (x) + 2 sin (x) - 3

1 + 2 sin (x) - 4 + 8 sin^{2} (x)

对同类项分组

= 2 sin (x) + 8 sin^{2} (x) + 1 - 4

数字相加/相减:

1 - 4 = - 3

= 8 sin^{2} (x) + 2 sin (x) - 3

= 8 sin^{2} (x) + 2 sin (x) - 3

= 8 sin^{2} (x) + 2 sin (x) - 3

- 3 + 2 sin (x) + 8 sin^{2} (x) = 0

用替代法求解

- 3 + 2 sin (x) + 8 sin^{2} (x) = 0

令

: sin (x) = u

- 3 + 2 u + 8 u^{2} = 0

- 3 + 2 u + 8 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{3}{4}

- 3 + 2 u + 8 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 8, b = 2, c = - 3

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

2^{2} - 4 \cdot 8 (- 3) = 10

2^{2} - 4 \cdot 8 (- 3)

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 8 \cdot 3

数字相乘:

4 \cdot 8 \cdot 3 = 96

= 2^{2} + 96

2^{2} = 4

= 4 + 96

数字相加:

4 + 96 = 100

= 100

因式分解数字:

100 = 1 0^{2}

= 1 0^{2}

使用根式运算法则:

n a^{n} = a

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- 2 + 10}{2 \cdot 8}, u_{2} = \frac{- 2 - 10}{2 \cdot 8}

u = \frac{- 2 + 10}{2 \cdot 8} : \frac{1}{2}

\frac{- 2 + 10}{2 \cdot 8}

数字相加/相减:

- 2 + 10 = 8

= \frac{8}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{8}{16}

约分:

8

= \frac{1}{2}

u = \frac{- 2 - 10}{2 \cdot 8} : - \frac{3}{4}

\frac{- 2 - 10}{2 \cdot 8}

数字相减:

- 2 - 10 = - 12

= \frac{- 12}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 12}{16}

使用分式法则:

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

= - \frac{12}{16}

约分:

4

= - \frac{3}{4}

二次方程组的解是:

u = \frac{1}{2}, u = - \frac{3}{4}

u = sin (x)

代回

sin (x) = \frac{1}{2}, sin (x) = - \frac{3}{4}

sin (x) = \frac{1}{2}, sin (x) = - \frac{3}{4}

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = - \frac{3}{4} : x = arcsin (- \frac{3}{4}) + 2 πn, x = π + arcsin (\frac{3}{4}) + 2 πn

sin (x) = - \frac{3}{4}

使用反三角函数性质

sin (x) = - \frac{3}{4}

sin (x) = - \frac{3}{4}

的通解

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

x = arcsin (- \frac{3}{4}) + 2 πn, x = π + arcsin (\frac{3}{4}) + 2 πn

x = arcsin (- \frac{3}{4}) + 2 πn, x = π + arcsin (\frac{3}{4}) + 2 πn

合并所有解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = arcsin (- \frac{3}{4}) + 2 πn, x = π + arcsin (\frac{3}{4}) + 2 πn

以小数形式表示解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = - 0.84806 \dots + 2 πn, x = π + 0.84806 \dots + 2 πn

作图

流行的例子

4 cos^{4} (x) - 1 = 0

arccos(2x)-arccos(x)= pi/3

arccos (2 x) - arccos (x) = \frac{π}{3}

cos(2θ)= 1/2 ,0<= θ<= 2pi

cos (2 θ) = \frac{1}{2}, 0 \leq θ \leq 2 π

tan(x-pi/2)=(sqrt(3))/3

tan (x - \frac{π}{2}) = \frac{3}{3}

2 = 2 cos (3 x)