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受欢迎的三角函数 >

sin^2(x)+2cos(x)=0

解答

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

解答

x = 1.99787 \dots + 2 πn, x = - 1.99787 \dots + 2 πn

+1

度数

x = 114.46980 \dots^{\circ} + 36 0^{\circ} n, x = - 114.46980 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

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

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

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

将解分隔开

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

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

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

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 1 = 2

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

使用分式法则:

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

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

消掉

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

\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

= - (2 - 1)

打开括号

= - (- 1) - (2)

使用加减运算法则

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

= 1 - 2

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

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

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 1 = 2

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

使用分式法则:

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

- 2 - 22 = - (2 + 22)

= \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

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 = - 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 = - arccos (- a) + 2 πn

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

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

以小数形式表示解

x = 1.99787 \dots + 2 πn, x = - 1.99787 \dots + 2 πn

作图

流行的例子

sin (a) = 0.36

sin (a) = 0.35

sin(2x)=cos(2x)+1,\forall 0<= θ<2pi

sin (2 x) = cos (2 x) + 1, \forall 0 \leq θ < 2 π

sin (x) = \frac{5}{14}

tan(2θ)=-(280}{\frac{(0-0))/2}

tan (2 θ) = - \frac{280}{\frac{( 0 - 0 )}{2}}