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受欢迎的三角函数 >

5-7sin(A)-2cos^2(A)=0

解答

5 - 7 sin (A) - 2 cos^{2} (A) = 0

解答

A = \frac{π}{6} + 2 πn, A = \frac{5 π}{6} + 2 πn

+1

度数

A = 3 0^{\circ} + 36 0^{\circ} n, A = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

5 - 7 sin (A) - 2 cos^{2} (A) = 0

使用三角恒等式改写

5 - 2 cos^{2} (A) - 7 sin (A)

使用毕达哥拉斯恒等式:

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

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

= 5 - 2 (1 - sin^{2} (A)) - 7 sin (A)

化简

5 - 2 (1 - sin^{2} (A)) - 7 sin (A) : 2 sin^{2} (A) - 7 sin (A) + 3

5 - 2 (1 - sin^{2} (A)) - 7 sin (A)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = - 2, b = 1, c = sin^{2} (A)

= - 2 \cdot 1 - (- 2) sin^{2} (A)

使用加减运算法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= - 2 + 2 sin^{2} (A)

= 5 - 2 + 2 sin^{2} (A) - 7 sin (A)

数字相减:

5 - 2 = 3

= 2 sin^{2} (A) - 7 sin (A) + 3

= 2 sin^{2} (A) - 7 sin (A) + 3

3 + 2 sin^{2} (A) - 7 sin (A) = 0

用替代法求解

3 + 2 sin^{2} (A) - 7 sin (A) = 0

令

: sin (A) = u

3 + 2 u^{2} - 7 u = 0

3 + 2 u^{2} - 7 u = 0 : u = 3, u = \frac{1}{2}

3 + 2 u^{2} - 7 u = 0

改写成标准形式

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

2 u^{2} - 7 u + 3 = 0

使用求根公式求解

2 u^{2} - 7 u + 3 = 0

二次方程求根公式:

若

a = 2, b = - 7, c = 3

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 3 = 5

(- 7)^{2} - 4 \cdot 2 \cdot 3

使用指数法则:

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

若

n

是偶数

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

= 7^{2} - 4 \cdot 2 \cdot 3

数字相乘:

4 \cdot 2 \cdot 3 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

数字相减:

49 - 24 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 7 ) \pm 5}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- ( - 7 ) + 5}{2 \cdot 2}, u_{2} = \frac{- ( - 7 ) - 5}{2 \cdot 2}

u = \frac{- ( - 7 ) + 5}{2 \cdot 2} : 3

\frac{- ( - 7 ) + 5}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{7 + 5}{2 \cdot 2}

数字相加:

7 + 5 = 12

= \frac{12}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{12}{4}

数字相除:

\frac{12}{4} = 3

= 3

u = \frac{- ( - 7 ) - 5}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 7 ) - 5}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{7 - 5}{2 \cdot 2}

数字相减:

7 - 5 = 2

= \frac{2}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

u = 3, u = \frac{1}{2}

u = sin (A)

代回

sin (A) = 3, sin (A) = \frac{1}{2}

sin (A) = 3, sin (A) = \frac{1}{2}

sin (A) = 3 :

无解

sin (A) = 3

- 1 \leq sin (x) \leq 1

无解

sin (A) = \frac{1}{2} : A = \frac{π}{6} + 2 πn, A = \frac{5 π}{6} + 2 πn

sin (A) = \frac{1}{2}

sin (A) = \frac{1}{2}

的通解

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

A = \frac{π}{6} + 2 πn, A = \frac{5 π}{6} + 2 πn

A = \frac{π}{6} + 2 πn, A = \frac{5 π}{6} + 2 πn

合并所有解

A = \frac{π}{6} + 2 πn, A = \frac{5 π}{6} + 2 πn

作图

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