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受欢迎的三角函数 >

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

解答

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

解答

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

+1

度数

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = - 13 5^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

\frac{1}{2}

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

化简

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

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

将项转换为分式:

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

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

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

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

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

使用三角恒等式改写

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

使用倍角公式:

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

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

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

化简

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = 2, b = cos (2 x), c = cos^{2} (x)

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

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

同类项相加:

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

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

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

1 - 2 cos^{2} (x) = 0

用替代法求解

1 - 2 cos^{2} (x) = 0

令

: cos (x) = u

1 - 2 u^{2} = 0

1 - 2 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

1 - 2 u^{2} = 0

将

1

到右边

1 - 2 u^{2} = 0

两边减去

1

1 - 2 u^{2} - 1 = 0 - 1

化简

- 2 u^{2} = - 1

- 2 u^{2} = - 1

两边除以

- 2

- 2 u^{2} = - 1

两边除以

- 2

\frac{- 2 u ^{2}}{- 2} = \frac{- 1}{- 2}

化简

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

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

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

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

u = cos (x)

代回

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

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

cos (x) = \frac{1}{2} : x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

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

使用反三角函数性质

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

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

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (x) = - \frac{1}{2} : x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

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

使用反三角函数性质

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

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

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

合并所有解

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn, x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

以小数形式表示解

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

作图

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