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受欢迎的三角函数 >

6-9sin(θ)-4cos^2(θ)=0

解答

6 - 9 sin (θ) - 4 cos^{2} (θ) = 0

解答

θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn

+1

度数

θ = 14.47751 \dots^{\circ} + 36 0^{\circ} n, θ = 165.52248 \dots^{\circ} + 36 0^{\circ} n

求解步骤

6 - 9 sin (θ) - 4 cos^{2} (θ) = 0

使用三角恒等式改写

6 - 4 cos^{2} (θ) - 9 sin (θ)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

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

= 6 - 4 (1 - sin^{2} (θ)) - 9 sin (θ)

化简

6 - 4 (1 - sin^{2} (θ)) - 9 sin (θ) : 4 sin^{2} (θ) - 9 sin (θ) + 2

6 - 4 (1 - sin^{2} (θ)) - 9 sin (θ)

乘开

- 4 (1 - sin^{2} (θ)) : - 4 + 4 sin^{2} (θ)

- 4 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (θ)

= - 4 \cdot 1 - (- 4) sin^{2} (θ)

使用加减运算法则

- (- a) = a

= - 4 \cdot 1 + 4 sin^{2} (θ)

数字相乘:

4 \cdot 1 = 4

= - 4 + 4 sin^{2} (θ)

= 6 - 4 + 4 sin^{2} (θ) - 9 sin (θ)

数字相减:

6 - 4 = 2

= 4 sin^{2} (θ) - 9 sin (θ) + 2

= 4 sin^{2} (θ) - 9 sin (θ) + 2

2 + 4 sin^{2} (θ) - 9 sin (θ) = 0

用替代法求解

2 + 4 sin^{2} (θ) - 9 sin (θ) = 0

令

: sin (θ) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 4 \cdot 2 = 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 \cdot 4}

将解分隔开

u_{1} = \frac{- ( - 9 ) + 7}{2 \cdot 4}, u_{2} = \frac{- ( - 9 ) - 7}{2 \cdot 4}

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

\frac{- ( - 9 ) + 7}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{9 + 7}{2 \cdot 4}

数字相加:

9 + 7 = 16

= \frac{16}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{16}{8}

数字相除:

\frac{16}{8} = 2

= 2

u = \frac{- ( - 9 ) - 7}{2 \cdot 4} : \frac{1}{4}

\frac{- ( - 9 ) - 7}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{9 - 7}{2 \cdot 4}

数字相减:

9 - 7 = 2

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

数字相乘:

2 \cdot 4 = 8

= \frac{2}{8}

约分:

2

= \frac{1}{4}

二次方程组的解是:

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

u = sin (θ)

代回

sin (θ) = 2, sin (θ) = \frac{1}{4}

sin (θ) = 2, sin (θ) = \frac{1}{4}

sin (θ) = 2 :

无解

sin (θ) = 2

- 1 \leq sin (x) \leq 1

无解

sin (θ) = \frac{1}{4} : θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

sin (θ) = \frac{1}{4}

使用反三角函数性质

sin (θ) = \frac{1}{4}

sin (θ) = \frac{1}{4}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

合并所有解

θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

以小数形式表示解

θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn

作图

流行的例子

2cos^2(x)-1cos(x)=0

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sec (x) = - 1.5

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cos (a) = \frac{9}{14}

\frac{π}{2} = 2 sin (x)