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受欢迎的三角函数 >

2sin^2(x)+9cos(x)=6

解答

2 sin^{2} (x) + 9 cos (x) = 6

解答

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

+1

度数

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

求解步骤

2 sin^{2} (x) + 9 cos (x) = 6

两边减去

6

2 sin^{2} (x) + 9 cos (x) - 6 = 0

使用三角恒等式改写

- 6 + 2 sin^{2} (x) + 9 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)

= - 6 + 2 (1 - cos^{2} (x)) + 9 cos (x)

化简

- 6 + 2 (1 - cos^{2} (x)) + 9 cos (x) : 9 cos (x) - 2 cos^{2} (x) - 4

- 6 + 2 (1 - cos^{2} (x)) + 9 cos (x)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (x)

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

数字相乘:

2 \cdot 1 = 2

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

= - 6 + 2 - 2 cos^{2} (x) + 9 cos (x)

数字相加/相减:

- 6 + 2 = - 4

= 9 cos (x) - 2 cos^{2} (x) - 4

= 9 cos (x) - 2 cos^{2} (x) - 4

- 4 - 2 cos^{2} (x) + 9 cos (x) = 0

用替代法求解

- 4 - 2 cos^{2} (x) + 9 cos (x) = 0

令

: cos (x) = u

- 4 - 2 u^{2} + 9 u = 0

- 4 - 2 u^{2} + 9 u = 0 : u = \frac{1}{2}, u = 4

- 4 - 2 u^{2} + 9 u = 0

改写成标准形式

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

- 2 u^{2} + 9 u - 4 = 0

使用求根公式求解

- 2 u^{2} + 9 u - 4 = 0

二次方程求根公式:

若

a = - 2, b = 9, c = - 4

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 ( - 2 ) ( - 4 )}{2 ( - 2 )}

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 ( - 2 ) ( - 4 )}{2 ( - 2 )}

9^{2} - 4 (- 2) (- 4) = 7

9^{2} - 4 (- 2) (- 4)

使用法则

- (- a) = a

= 9^{2} - 4 \cdot 2 \cdot 4

数字相乘:

4 \cdot 2 \cdot 4 = 32

= 9^{2} - 32

9^{2} = 81

= 81 - 32

数字相减:

81 - 32 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

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

将解分隔开

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

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

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

去除括号:

(- a) = - a

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

数字相加/相减:

- 9 + 7 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

使用分式法则:

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

= \frac{2}{4}

约分:

2

= \frac{1}{2}

u = \frac{- 9 - 7}{2 ( - 2 )} : 4

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

去除括号:

(- a) = - a

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

数字相减:

- 9 - 7 = - 16

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 16}{- 4}

使用分式法则:

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

= \frac{16}{4}

数字相除:

\frac{16}{4} = 4

= 4

二次方程组的解是:

u = \frac{1}{2}, u = 4

u = cos (x)

代回

cos (x) = \frac{1}{2}, cos (x) = 4

cos (x) = \frac{1}{2}, cos (x) = 4

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

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

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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = 4 :

无解

cos (x) = 4

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

作图

流行的例子

sqrt(3)sin(x)+cos(x)=0

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2 cos^{2} (x) - 3 sin (x) = 0

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3 cos (x) = 1 + cos (x)