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受欢迎的三角函数 >

2cos(x)=sin^2(x)

解答

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

解答

x = 1.14371 \dots + 2 πn, x = 2 π - 1.14371 \dots + 2 πn

+1

度数

x = 65.53019 \dots^{\circ} + 36 0^{\circ} n, x = 294.46980 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

sin^{2} (x)

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

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

打开括号

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

使用加减运算法则

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

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

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

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

用替代法求解

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

令

: cos (x) = u

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

- 1 + u^{2} + 2 u = 0 : u = - 1 + 2, u = - 1 - 2

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 1, b = 2, c = - 1

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

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

2^{2} - 4 \cdot 1 \cdot (- 1) = 22

2^{2} - 4 \cdot 1 \cdot (- 1)

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

数字相加:

4 + 4 = 8

= 8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

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

将解分隔开

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

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

\frac{- 2 + 2 2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

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

分解

- 2 + 22 : 2 (- 1 + 2)

- 2 + 22

改写为

= - 2 \cdot 1 + 22

因式分解出通项

2

= 2 (- 1 + 2)

= \frac{2 ( - 1 + 2 )}{2}

数字相除:

\frac{2}{2} = 1

= - 1 + 2

u = \frac{- 2 - 2 2}{2 \cdot 1} : - 1 - 2

\frac{- 2 - 2 2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

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

分解

- 2 - 22 : - 2 (1 + 2)

- 2 - 22

改写为

= - 2 \cdot 1 - 22

因式分解出通项

2

= - 2 (1 + 2)

= - \frac{2 ( 1 + 2 )}{2}

数字相除:

\frac{2}{2} = 1

= - (1 + 2)

取负

- (1 + 2) = - 1 - 2

= - 1 - 2

二次方程组的解是:

u = - 1 + 2, u = - 1 - 2

u = cos (x)

代回

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

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

cos (x) = - 1 + 2 : x = arccos (- 1 + 2) + 2 πn, x = 2 π - arccos (- 1 + 2) + 2 πn

cos (x) = - 1 + 2

使用反三角函数性质

cos (x) = - 1 + 2

cos (x) = - 1 + 2

的通解

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

x = arccos (- 1 + 2) + 2 πn, x = 2 π - arccos (- 1 + 2) + 2 πn

x = arccos (- 1 + 2) + 2 πn, x = 2 π - arccos (- 1 + 2) + 2 πn

cos (x) = - 1 - 2 :

无解

cos (x) = - 1 - 2

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = arccos (- 1 + 2) + 2 πn, x = 2 π - arccos (- 1 + 2) + 2 πn

以小数形式表示解

x = 1.14371 \dots + 2 πn, x = 2 π - 1.14371 \dots + 2 πn

作图

流行的例子

-3sin(x/2)=1,-pi/2 <= x<= pi/2

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2 sin^{2} (θ) - 5 sin (θ) - 3 = 0

10=6.7+13.75tan(θ)-4.718(1+tan^2(θ))

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