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受欢迎的三角函数 >

4sin^2(θ)-8cos(θ)+1=0

解答

4 sin^{2} (θ) - 8 cos (θ) + 1 = 0

解答

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

+1

度数

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

求解步骤

4 sin^{2} (θ) - 8 cos (θ) + 1 = 0

使用三角恒等式改写

1 + 4 sin^{2} (θ) - 8 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)

= 1 + 4 (1 - cos^{2} (θ)) - 8 cos (θ)

化简

1 + 4 (1 - cos^{2} (θ)) - 8 cos (θ) : - 4 cos^{2} (θ) - 8 cos (θ) + 5

1 + 4 (1 - cos^{2} (θ)) - 8 cos (θ)

乘开

4 (1 - cos^{2} (θ)) : 4 - 4 cos^{2} (θ)

4 (1 - cos^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (θ)

= 4 \cdot 1 - 4 cos^{2} (θ)

数字相乘:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (θ)

= 1 + 4 - 4 cos^{2} (θ) - 8 cos (θ)

数字相加:

1 + 4 = 5

= - 4 cos^{2} (θ) - 8 cos (θ) + 5

= - 4 cos^{2} (θ) - 8 cos (θ) + 5

5 - 4 cos^{2} (θ) - 8 cos (θ) = 0

用替代法求解

5 - 4 cos^{2} (θ) - 8 cos (θ) = 0

令

: cos (θ) = u

5 - 4 u^{2} - 8 u = 0

5 - 4 u^{2} - 8 u = 0 : u = - \frac{5}{2}, u = \frac{1}{2}

5 - 4 u^{2} - 8 u = 0

改写成标准形式

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

- 4 u^{2} - 8 u + 5 = 0

使用求根公式求解

- 4 u^{2} - 8 u + 5 = 0

二次方程求根公式:

若

a = - 4, b = - 8, c = 5

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

(- 8)^{2} - 4 (- 4) \cdot 5 = 12

(- 8)^{2} - 4 (- 4) \cdot 5

使用法则

- (- a) = a

= (- 8)^{2} + 4 \cdot 4 \cdot 5

使用指数法则:

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

若

n

是偶数

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 4 \cdot 5

数字相乘:

4 \cdot 4 \cdot 5 = 80

= 8^{2} + 80

8^{2} = 64

= 64 + 80

数字相加:

64 + 80 = 144

= 144

因式分解数字:

144 = 1 2^{2}

= 1 2^{2}

使用根式运算法则:

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- ( - 8 ) \pm 12}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- ( - 8 ) + 12}{2 ( - 4 )}, u_{2} = \frac{- ( - 8 ) - 12}{2 ( - 4 )}

u = \frac{- ( - 8 ) + 12}{2 ( - 4 )} : - \frac{5}{2}

\frac{- ( - 8 ) + 12}{2 ( - 4 )}

去除括号:

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

= \frac{8 + 12}{- 2 \cdot 4}

数字相加:

8 + 12 = 20

= \frac{20}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{20}{- 8}

使用分式法则:

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

= - \frac{20}{8}

约分:

4

= - \frac{5}{2}

u = \frac{- ( - 8 ) - 12}{2 ( - 4 )} : \frac{1}{2}

\frac{- ( - 8 ) - 12}{2 ( - 4 )}

去除括号:

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

= \frac{8 - 12}{- 2 \cdot 4}

数字相减:

8 - 12 = - 4

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

使用分式法则:

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

= \frac{4}{8}

约分:

4

= \frac{1}{2}

二次方程组的解是:

u = - \frac{5}{2}, u = \frac{1}{2}

u = cos (θ)

代回

cos (θ) = - \frac{5}{2}, cos (θ) = \frac{1}{2}

cos (θ) = - \frac{5}{2}, cos (θ) = \frac{1}{2}

cos (θ) = - \frac{5}{2} :

无解

cos (θ) = - \frac{5}{2}

- 1 \leq cos (x) \leq 1

无解

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

合并所有解

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

作图

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