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受欢迎的三角函数 >

2sin^2(x)-3cos(x)=3

解答

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

解答

x = π + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

+1

度数

x = 18 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

3

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

使用三角恒等式改写

- 3 + 2 sin^{2} (x) - 3 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)

= - 3 + 2 (1 - cos^{2} (x)) - 3 cos (x)

化简

- 3 + 2 (1 - cos^{2} (x)) - 3 cos (x) : - 2 cos^{2} (x) - 3 cos (x) - 1

- 3 + 2 (1 - cos^{2} (x)) - 3 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)

= - 3 + 2 - 2 cos^{2} (x) - 3 cos (x)

数字相加/相减:

- 3 + 2 = - 1

= - 2 cos^{2} (x) - 3 cos (x) - 1

= - 2 cos^{2} (x) - 3 cos (x) - 1

- 1 - 2 cos^{2} (x) - 3 cos (x) = 0

用替代法求解

- 1 - 2 cos^{2} (x) - 3 cos (x) = 0

令

: cos (x) = u

- 1 - 2 u^{2} - 3 u = 0

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

- 1 - 2 u^{2} - 3 u = 0

改写成标准形式

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

- 2 u^{2} - 3 u - 1 = 0

使用求根公式求解

- 2 u^{2} - 3 u - 1 = 0

二次方程求根公式:

若

a = - 2, b = - 3, c = - 1

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

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

(- 3)^{2} - 4 (- 2) (- 1) = 1

(- 3)^{2} - 4 (- 2) (- 1)

使用法则

- (- a) = a

= (- 3)^{2} - 4 \cdot 2 \cdot 1

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数字相减:

9 - 8 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 1}{2 ( - 2 )}, u_{2} = \frac{- ( - 3 ) - 1}{2 ( - 2 )}

u = \frac{- ( - 3 ) + 1}{2 ( - 2 )} : - 1

\frac{- ( - 3 ) + 1}{2 ( - 2 )}

去除括号:

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

= \frac{3 + 1}{- 2 \cdot 2}

数字相加:

3 + 1 = 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{- 4}

使用分式法则:

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

= - \frac{4}{4}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 3 ) - 1}{2 ( - 2 )} : - \frac{1}{2}

\frac{- ( - 3 ) - 1}{2 ( - 2 )}

去除括号:

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

= \frac{3 - 1}{- 2 \cdot 2}

数字相减:

3 - 1 = 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}

二次方程组的解是:

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

u = cos (x)

代回

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

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

cos (x) = - 1 : x = π + 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 = π + 2 πn

x = π + 2 πn

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

合并所有解

x = π + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

作图

流行的例子

4cos(x)=3+8cos(x)

4 cos (x) = 3 + 8 cos (x)

sin^2(θ)-3sin(θ)=0

sin^{2} (θ) - 3 sin (θ) = 0

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

-3cos(2θ)+5=7sin(θ)+5

- 3 cos (2 θ) + 5 = 7 sin (θ) + 5

3+3cos(x)=2sin^2(x)

3 + 3 cos (x) = 2 sin^{2} (x)