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受欢迎的三角函数 >

-4sin(x)=cos^2(x)+1,0<= x<= 2pi

解答

- 4 sin (x) = cos^{2} (x) + 1, 0 \leq x \leq 2 π

解答

x = π + 0.46619 \dots, x = - 0.46619 \dots + 2 π

+1

度数

x = 206.71095 \dots^{\circ}, x = 333.28904 \dots^{\circ}

求解步骤

- 4 sin (x) = cos^{2} (x) + 1, 0 \leq x \leq 2 π

两边减去

cos^{2} (x) + 1

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

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

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

打开括号

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

使用加减运算法则

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

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

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

数字相减:

- 1 - 1 = - 2

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

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

- 2 + sin^{2} (x) - 4 sin (x) = 0

用替代法求解

- 2 + sin^{2} (x) - 4 sin (x) = 0

令

: sin (x) = u

- 2 + u^{2} - 4 u = 0

- 2 + u^{2} - 4 u = 0 : u = 2 + 6, u = 2 - 6

- 2 + u^{2} - 4 u = 0

改写成标准形式

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

u^{2} - 4 u - 2 = 0

使用求根公式求解

u^{2} - 4 u - 2 = 0

二次方程求根公式:

若

a = 1, b = - 4, c = - 2

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

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 2) = 26

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 4^{2} + 8

4^{2} = 16

= 16 + 8

数字相加:

16 + 8 = 24

= 24

24

质因数分解:

2^{3} \cdot 3

24

24

除以

224 = 12 \cdot 2

= 2 \cdot 12

12

除以

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 3

= 2^{3} \cdot 3

= 2^{3} \cdot 3

使用指数法则:

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

= 2^{2} \cdot 2 \cdot 3

使用根式运算法则:

n ab = n a n b

= 2^{2} 2 \cdot 3

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 2 \cdot 3

整理后得

= 26

u_{1, 2} = \frac{- ( - 4 ) \pm 2 6}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 4 ) + 2 6}{2 \cdot 1}, u_{2} = \frac{- ( - 4 ) - 2 6}{2 \cdot 1}

u = \frac{- ( - 4 ) + 2 6}{2 \cdot 1} : 2 + 6

\frac{- ( - 4 ) + 2 6}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{4 + 2 6}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4 + 2 6}{2}

分解

4 + 26 : 2 (2 + 6)

4 + 26

改写为

= 2 \cdot 2 + 26

因式分解出通项

2

= 2 (2 + 6)

= \frac{2 ( 2 + 6 )}{2}

数字相除:

\frac{2}{2} = 1

= 2 + 6

u = \frac{- ( - 4 ) - 2 6}{2 \cdot 1} : 2 - 6

\frac{- ( - 4 ) - 2 6}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{4 - 2 6}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4 - 2 6}{2}

分解

4 - 26 : 2 (2 - 6)

4 - 26

改写为

= 2 \cdot 2 - 26

因式分解出通项

2

= 2 (2 - 6)

= \frac{2 ( 2 - 6 )}{2}

数字相除:

\frac{2}{2} = 1

= 2 - 6

二次方程组的解是:

u = 2 + 6, u = 2 - 6

u = sin (x)

代回

sin (x) = 2 + 6, sin (x) = 2 - 6

sin (x) = 2 + 6, sin (x) = 2 - 6

sin (x) = 2 + 6, 0 \leq x \leq 2 π :

无解

sin (x) = 2 + 6, 0 \leq x \leq 2 π

- 1 \leq sin (x) \leq 1

无解

sin (x) = 2 - 6, 0 \leq x \leq 2 π : x = π + arcsin (6 - 2), x = arcsin (2 - 6) + 2 π

sin (x) = 2 - 6, 0 \leq x \leq 2 π

使用反三角函数性质

sin (x) = 2 - 6

sin (x) = 2 - 6

的通解

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

x = arcsin (2 - 6) + 2 πn, x = π + arcsin (- 2 + 6) + 2 πn

x = arcsin (2 - 6) + 2 πn, x = π + arcsin (- 2 + 6) + 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = π + arcsin (6 - 2), x = arcsin (2 - 6) + 2 π

合并所有解

x = π + arcsin (6 - 2), x = arcsin (2 - 6) + 2 π

以小数形式表示解

x = π + 0.46619 \dots, x = - 0.46619 \dots + 2 π

作图

流行的例子

tan(2x)cos(2x)+cot(2x)sin(2x)=1

tan (2 x) cos (2 x) + cot (2 x) sin (2 x) = 1

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

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

sin(x)=(1.2)/(sqrt(10))

sin (x) = \frac{1.2}{10}

solvefor t,x=-3cos(pit)

so l v e f or t, x = - 3 cos (π t)

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