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受欢迎的三角函数 >

3sin^2(θ)-6cos(θ)-3=-4cos(θ)

解答

3 sin^{2} (θ) - 6 cos (θ) - 3 = - 4 cos (θ)

解答

θ = 2.30052 \dots + 2 πn, θ = - 2.30052 \dots + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

+1

度数

θ = 131.81031 \dots^{\circ} + 36 0^{\circ} n, θ = - 131.81031 \dots^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

3 sin^{2} (θ) - 6 cos (θ) - 3 = - 4 cos (θ)

两边减去

- 4 cos (θ)

3 sin^{2} (θ) - 2 cos (θ) - 3 = 0

使用三角恒等式改写

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

化简

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

- 3 - 2 cos (θ) + 3 (1 - cos^{2} (θ))

乘开

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

3 (1 - cos^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = cos^{2} (θ)

= 3 \cdot 1 - 3 cos^{2} (θ)

数字相乘:

3 \cdot 1 = 3

= 3 - 3 cos^{2} (θ)

= - 3 - 2 cos (θ) + 3 - 3 cos^{2} (θ)

化简

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

- 3 - 2 cos (θ) + 3 - 3 cos^{2} (θ)

对同类项分组

= - 2 cos (θ) - 3 cos^{2} (θ) - 3 + 3

- 3 + 3 = 0

= - 3 cos^{2} (θ) - 2 cos (θ)

= - 3 cos^{2} (θ) - 2 cos (θ)

= - 3 cos^{2} (θ) - 2 cos (θ)

- 2 cos (θ) - 3 cos^{2} (θ) = 0

用替代法求解

- 2 cos (θ) - 3 cos^{2} (θ) = 0

令

: cos (θ) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 3, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 3 ) \cdot 0}{2 ( - 3 )}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 3 ) \cdot 0}{2 ( - 3 )}

(- 2)^{2} - 4 (- 3) \cdot 0 = 2

(- 2)^{2} - 4 (- 3) \cdot 0

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 3 \cdot 0

使用指数法则:

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

若

n

是偶数

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

= 2^{2} + 4 \cdot 3 \cdot 0

使用法则

0 \cdot a = 0

= 2^{2} + 0

2^{2} + 0 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 2

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

将解分隔开

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

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

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

去除括号:

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

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

数字相加:

2 + 2 = 4

= \frac{4}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{4}{- 6}

使用分式法则:

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

= - \frac{4}{6}

约分:

2

= - \frac{2}{3}

u = \frac{- ( - 2 ) - 2}{2 ( - 3 )} : 0

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

去除括号:

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

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

数字相减:

2 - 2 = 0

= \frac{0}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{0}{- 6}

使用分式法则:

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

= - \frac{0}{6}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

二次方程组的解是:

u = - \frac{2}{3}, u = 0

u = cos (θ)

代回

cos (θ) = - \frac{2}{3}, cos (θ) = 0

cos (θ) = - \frac{2}{3}, cos (θ) = 0

cos (θ) = - \frac{2}{3} : θ = arccos (- \frac{2}{3}) + 2 πn, θ = - arccos (- \frac{2}{3}) + 2 πn

cos (θ) = - \frac{2}{3}

使用反三角函数性质

cos (θ) = - \frac{2}{3}

cos (θ) = - \frac{2}{3}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- \frac{2}{3}) + 2 πn, θ = - arccos (- \frac{2}{3}) + 2 πn

θ = arccos (- \frac{2}{3}) + 2 πn, θ = - arccos (- \frac{2}{3}) + 2 πn

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

cos (θ) = 0

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}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

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

合并所有解

θ = arccos (- \frac{2}{3}) + 2 πn, θ = - arccos (- \frac{2}{3}) + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

以小数形式表示解

θ = 2.30052 \dots + 2 πn, θ = - 2.30052 \dots + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

作图

流行的例子

2 - sin (x) = 0

6tan^2(θ)-5tan(θ)+1=0

6 tan^{2} (θ) - 5 tan (θ) + 1 = 0

6 sin (x) - 24 = 0

-2sin(x)-2sin(2x)=0

- 2 sin (x) - 2 sin (2 x) = 0

((sin^2(x)+1))/(1-sin^2(x))-1=2

\frac{( sin ^{2} ( x ) + 1 )}{1 - sin ^{2} ( x )} - 1 = 2