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受欢迎的三角函数 >

6cos^2(θ)-5=6sin^2(θ)+sin(θ)

解答

6 cos^{2} (θ) - 5 = 6 sin^{2} (θ) + sin (θ)

解答

θ = - 0.33983 \dots + 2 πn, θ = π + 0.33983 \dots + 2 πn, θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn

+1

度数

θ = - 19.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 199.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 14.47751 \dots^{\circ} + 36 0^{\circ} n, θ = 165.52248 \dots^{\circ} + 36 0^{\circ} n

求解步骤

6 cos^{2} (θ) - 5 = 6 sin^{2} (θ) + sin (θ)

两边减去

6 sin^{2} (θ) + sin (θ)

6 cos^{2} (θ) - 5 - 6 sin^{2} (θ) - sin (θ) = 0

使用三角恒等式改写

- 5 - sin (θ) + 6 cos^{2} (θ) - 6 sin^{2} (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)

= - 5 - sin (θ) + 6 (1 - sin^{2} (θ)) - 6 sin^{2} (θ)

化简

- 5 - sin (θ) + 6 (1 - sin^{2} (θ)) - 6 sin^{2} (θ) : - 12 sin^{2} (θ) - sin (θ) + 1

- 5 - sin (θ) + 6 (1 - sin^{2} (θ)) - 6 sin^{2} (θ)

乘开

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

6 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 6, b = 1, c = sin^{2} (θ)

= 6 \cdot 1 - 6 sin^{2} (θ)

数字相乘:

6 \cdot 1 = 6

= 6 - 6 sin^{2} (θ)

= - 5 - sin (θ) + 6 - 6 sin^{2} (θ) - 6 sin^{2} (θ)

化简

- 5 - sin (θ) + 6 - 6 sin^{2} (θ) - 6 sin^{2} (θ) : - 12 sin^{2} (θ) - sin (θ) + 1

- 5 - sin (θ) + 6 - 6 sin^{2} (θ) - 6 sin^{2} (θ)

同类项相加:

- 6 sin^{2} (θ) - 6 sin^{2} (θ) = - 12 sin^{2} (θ)

= - 5 - sin (θ) + 6 - 12 sin^{2} (θ)

对同类项分组

= - sin (θ) - 12 sin^{2} (θ) - 5 + 6

数字相加/相减:

- 5 + 6 = 1

= - 12 sin^{2} (θ) - sin (θ) + 1

= - 12 sin^{2} (θ) - sin (θ) + 1

= - 12 sin^{2} (θ) - sin (θ) + 1

1 - sin (θ) - 12 sin^{2} (θ) = 0

用替代法求解

1 - sin (θ) - 12 sin^{2} (θ) = 0

令

: sin (θ) = u

1 - u - 12 u^{2} = 0

1 - u - 12 u^{2} = 0 : u = - \frac{1}{3}, u = \frac{1}{4}

1 - u - 12 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 12, b = - 1, c = 1

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 12 ) \cdot 1}{2 ( - 12 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 12 ) \cdot 1}{2 ( - 12 )}

(- 1)^{2} - 4 (- 12) \cdot 1 = 7

(- 1)^{2} - 4 (- 12) \cdot 1

使用法则

- (- a) = a

= (- 1)^{2} + 4 \cdot 12 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 12 \cdot 1 = 48

4 \cdot 12 \cdot 1

数字相乘:

4 \cdot 12 \cdot 1 = 48

= 48

= 1 + 48

数字相加:

1 + 48 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 ( - 12 )}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 7}{2 ( - 12 )}, u_{2} = \frac{- ( - 1 ) - 7}{2 ( - 12 )}

u = \frac{- ( - 1 ) + 7}{2 ( - 12 )} : - \frac{1}{3}

\frac{- ( - 1 ) + 7}{2 ( - 12 )}

去除括号:

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

= \frac{1 + 7}{- 2 \cdot 12}

数字相加:

1 + 7 = 8

= \frac{8}{- 2 \cdot 12}

数字相乘:

2 \cdot 12 = 24

= \frac{8}{- 24}

使用分式法则:

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

= - \frac{8}{24}

约分:

8

= - \frac{1}{3}

u = \frac{- ( - 1 ) - 7}{2 ( - 12 )} : \frac{1}{4}

\frac{- ( - 1 ) - 7}{2 ( - 12 )}

去除括号:

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

= \frac{1 - 7}{- 2 \cdot 12}

数字相减:

1 - 7 = - 6

= \frac{- 6}{- 2 \cdot 12}

数字相乘:

2 \cdot 12 = 24

= \frac{- 6}{- 24}

使用分式法则:

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

= \frac{6}{24}

约分:

6

= \frac{1}{4}

二次方程组的解是:

u = - \frac{1}{3}, u = \frac{1}{4}

u = sin (θ)

代回

sin (θ) = - \frac{1}{3}, sin (θ) = \frac{1}{4}

sin (θ) = - \frac{1}{3}, sin (θ) = \frac{1}{4}

sin (θ) = - \frac{1}{3} : θ = arcsin (- \frac{1}{3}) + 2 πn, θ = π + arcsin (\frac{1}{3}) + 2 πn

sin (θ) = - \frac{1}{3}

使用反三角函数性质

sin (θ) = - \frac{1}{3}

sin (θ) = - \frac{1}{3}

的通解

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

θ = arcsin (- \frac{1}{3}) + 2 πn, θ = π + arcsin (\frac{1}{3}) + 2 πn

θ = arcsin (- \frac{1}{3}) + 2 πn, θ = π + arcsin (\frac{1}{3}) + 2 πn

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}{3}) + 2 πn, θ = π + arcsin (\frac{1}{3}) + 2 πn, θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

以小数形式表示解

θ = - 0.33983 \dots + 2 πn, θ = π + 0.33983 \dots + 2 πn, θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn

作图

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