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受欢迎的三角函数 >

2sin^2(θ)+3cos(θ)=3

解答

2 sin^{2} (θ) + 3 cos (θ) = 3

解答

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

+1

度数

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

求解步骤

2 sin^{2} (θ) + 3 cos (θ) = 3

两边减去

3

2 sin^{2} (θ) + 3 cos (θ) - 3 = 0

使用三角恒等式改写

- 3 + 2 sin^{2} (θ) + 3 cos (θ)

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} (θ)) + 3 cos (θ)

化简

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

- 3 + 2 (1 - cos^{2} (θ)) + 3 cos (θ)

乘开

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

2 (1 - cos^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (θ)

= 2 \cdot 1 - 2 cos^{2} (θ)

数字相乘:

2 \cdot 1 = 2

= 2 - 2 cos^{2} (θ)

= - 3 + 2 - 2 cos^{2} (θ) + 3 cos (θ)

数字相加/相减:

- 3 + 2 = - 1

= 3 cos (θ) - 2 cos^{2} (θ) - 1

= 3 cos (θ) - 2 cos^{2} (θ) - 1

- 1 - 2 cos^{2} (θ) + 3 cos (θ) = 0

用替代法求解

- 1 - 2 cos^{2} (θ) + 3 cos (θ) = 0

令

: cos (θ) = u

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

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

- 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

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 )} : \frac{1}{2}

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

去除括号:

(- a) = - a

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

数字相加/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}

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

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

去除括号:

(- a) = - a

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

数字相减:

- 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

二次方程组的解是:

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

u = cos (θ)

代回

cos (θ) = \frac{1}{2}, cos (θ) = 1

cos (θ) = \frac{1}{2}, cos (θ) = 1

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

cos (θ) = \frac{1}{2}

cos (θ) = \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}

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

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

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

cos (θ) = 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}

θ = 0 + 2 πn

θ = 0 + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

合并所有解

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

作图

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