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受欢迎的三角函数 >

2sin^2(x)+11cos(x)=7

解答

2 sin^{2} (x) + 11 cos (x) = 7

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) + 11 cos (x) = 7

两边减去

7

2 sin^{2} (x) + 11 cos (x) - 7 = 0

使用三角恒等式改写

- 7 + 11 cos (x) + 2 sin^{2} (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 + 11 cos (x) + 2 (1 - cos^{2} (x))

化简

- 7 + 11 cos (x) + 2 (1 - cos^{2} (x)) : 11 cos (x) - 2 cos^{2} (x) - 5

- 7 + 11 cos (x) + 2 (1 - cos^{2} (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)

= - 7 + 11 cos (x) + 2 - 2 cos^{2} (x)

化简

- 7 + 11 cos (x) + 2 - 2 cos^{2} (x) : 11 cos (x) - 2 cos^{2} (x) - 5

- 7 + 11 cos (x) + 2 - 2 cos^{2} (x)

对同类项分组

= 11 cos (x) - 2 cos^{2} (x) - 7 + 2

数字相加/相减:

- 7 + 2 = - 5

= 11 cos (x) - 2 cos^{2} (x) - 5

= 11 cos (x) - 2 cos^{2} (x) - 5

= 11 cos (x) - 2 cos^{2} (x) - 5

- 5 + 11 cos (x) - 2 cos^{2} (x) = 0

用替代法求解

- 5 + 11 cos (x) - 2 cos^{2} (x) = 0

令

: cos (x) = u

- 5 + 11 u - 2 u^{2} = 0

- 5 + 11 u - 2 u^{2} = 0 : u = \frac{1}{2}, u = 5

- 5 + 11 u - 2 u^{2} = 0

改写成标准形式

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

- 2 u^{2} + 11 u - 5 = 0

使用求根公式求解

- 2 u^{2} + 11 u - 5 = 0

二次方程求根公式:

若

a = - 2, b = 11, c = - 5

u_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 ( - 2 ) ( - 5 )}{2 ( - 2 )}

u_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 ( - 2 ) ( - 5 )}{2 ( - 2 )}

1 1^{2} - 4 (- 2) (- 5) = 9

1 1^{2} - 4 (- 2) (- 5)

使用法则

- (- a) = a

= 1 1^{2} - 4 \cdot 2 \cdot 5

数字相乘:

4 \cdot 2 \cdot 5 = 40

= 1 1^{2} - 40

1 1^{2} = 121

= 121 - 40

数字相减:

121 - 40 = 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{- 11 \pm 9}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 11 + 9}{2 ( - 2 )}, u_{2} = \frac{- 11 - 9}{2 ( - 2 )}

u = \frac{- 11 + 9}{2 ( - 2 )} : \frac{1}{2}

\frac{- 11 + 9}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 11 + 9}{- 2 \cdot 2}

数字相加/相减:

- 11 + 9 = - 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{- 11 - 9}{2 ( - 2 )} : 5

\frac{- 11 - 9}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 11 - 9}{- 2 \cdot 2}

数字相减:

- 11 - 9 = - 20

= \frac{- 20}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 20}{- 4}

使用分式法则:

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

= \frac{20}{4}

数字相除:

\frac{20}{4} = 5

= 5

二次方程组的解是:

u = \frac{1}{2}, u = 5

u = cos (x)

代回

cos (x) = \frac{1}{2}, cos (x) = 5

cos (x) = \frac{1}{2}, cos (x) = 5

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) = 5 :

无解

cos (x) = 5

- 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

sin (A) = \frac{5}{13}

2cos(x)=-sqrt(2)

2 cos (x) = - 2

3tan(3x)=sqrt(3)

3 tan (3 x) = 3

cos(2/3 x-pi/2)-1=0

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

sin (x) = - \frac{3}{2}