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受欢迎的三角函数 >

sin^2(x)-7cos(x)+7=0

解答

sin^{2} (x) - 7 cos (x) + 7 = 0

解答

x = 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n

求解步骤

sin^{2} (x) - 7 cos (x) + 7 = 0

使用三角恒等式改写

7 + sin^{2} (x) - 7 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)

= 7 + 1 - cos^{2} (x) - 7 cos (x)

化简

= - cos^{2} (x) - 7 cos (x) + 8

8 - cos^{2} (x) - 7 cos (x) = 0

用替代法求解

8 - cos^{2} (x) - 7 cos (x) = 0

令

: cos (x) = u

8 - u^{2} - 7 u = 0

8 - u^{2} - 7 u = 0 : u = - 8, u = 1

8 - u^{2} - 7 u = 0

改写成标准形式

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

- u^{2} - 7 u + 8 = 0

使用求根公式求解

- u^{2} - 7 u + 8 = 0

二次方程求根公式:

若

a = - 1, b = - 7, c = 8

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 1 ) \cdot 8}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 1 ) \cdot 8}{2 ( - 1 )}

(- 7)^{2} - 4 (- 1) \cdot 8 = 9

(- 7)^{2} - 4 (- 1) \cdot 8

使用法则

- (- a) = a

= (- 7)^{2} + 4 \cdot 1 \cdot 8

使用指数法则:

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

若

n

是偶数

(- 7)^{2} = 7^{2}

= 7^{2} + 4 \cdot 1 \cdot 8

数字相乘:

4 \cdot 1 \cdot 8 = 32

= 7^{2} + 32

7^{2} = 49

= 49 + 32

数字相加:

49 + 32 = 81

= 81

因式分解数字:

81 = 9^{2}

= 9^{2}

使用根式运算法则:

n a^{n} = a

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 7 ) \pm 9}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- ( - 7 ) + 9}{2 ( - 1 )}, u_{2} = \frac{- ( - 7 ) - 9}{2 ( - 1 )}

u = \frac{- ( - 7 ) + 9}{2 ( - 1 )} : - 8

\frac{- ( - 7 ) + 9}{2 ( - 1 )}

去除括号:

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

= \frac{7 + 9}{- 2 \cdot 1}

数字相加:

7 + 9 = 16

= \frac{16}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{16}{- 2}

使用分式法则:

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

= - \frac{16}{2}

数字相除:

\frac{16}{2} = 8

= - 8

u = \frac{- ( - 7 ) - 9}{2 ( - 1 )} : 1

\frac{- ( - 7 ) - 9}{2 ( - 1 )}

去除括号:

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

= \frac{7 - 9}{- 2 \cdot 1}

数字相减:

7 - 9 = - 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

使用分式法则:

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

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = - 8, u = 1

u = cos (x)

代回

cos (x) = - 8, cos (x) = 1

cos (x) = - 8, cos (x) = 1

cos (x) = - 8 :

无解

cos (x) = - 8

- 1 \leq cos (x) \leq 1

无解

cos (x) = 1 : x = 2 πn

cos (x) = 1

cos (x) = 1

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

合并所有解

x = 2 πn

作图

流行的例子

sin(x)-sqrt(3)=-sin(x)

sin (x) - 3 = - sin (x)

6 sin (θ) - 6 = 0

tan(5x)=sqrt(3)

tan (5 x) = 3

cos(2θ)=3cos(θ)-2

cos (2 θ) = 3 cos (θ) - 2

sin (x) = - \frac{5}{13}