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受欢迎的三角函数 >

2sin^2(θ)-5cos(θ)+1=0

解答

2 sin^{2} (θ) - 5 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

2 sin^{2} (θ) - 5 cos (θ) + 1 = 0

使用三角恒等式改写

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

化简

1 + 2 (1 - cos^{2} (θ)) - 5 cos (θ) : - 2 cos^{2} (θ) - 5 cos (θ) + 3

1 + 2 (1 - cos^{2} (θ)) - 5 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)

= 1 + 2 - 2 cos^{2} (θ) - 5 cos (θ)

数字相加:

1 + 2 = 3

= - 2 cos^{2} (θ) - 5 cos (θ) + 3

= - 2 cos^{2} (θ) - 5 cos (θ) + 3

3 - 2 cos^{2} (θ) - 5 cos (θ) = 0

用替代法求解

3 - 2 cos^{2} (θ) - 5 cos (θ) = 0

令

: cos (θ) = u

3 - 2 u^{2} - 5 u = 0

3 - 2 u^{2} - 5 u = 0 : u = - 3, u = \frac{1}{2}

3 - 2 u^{2} - 5 u = 0

改写成标准形式

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

- 2 u^{2} - 5 u + 3 = 0

使用求根公式求解

- 2 u^{2} - 5 u + 3 = 0

二次方程求根公式:

若

a = - 2, b = - 5, c = 3

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

(- 5)^{2} - 4 (- 2) \cdot 3 = 7

(- 5)^{2} - 4 (- 2) \cdot 3

使用法则

- (- a) = a

= (- 5)^{2} + 4 \cdot 2 \cdot 3

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 2 \cdot 3

数字相乘:

4 \cdot 2 \cdot 3 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

数字相加:

25 + 24 = 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{- ( - 5 ) \pm 7}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 7}{2 ( - 2 )}, u_{2} = \frac{- ( - 5 ) - 7}{2 ( - 2 )}

u = \frac{- ( - 5 ) + 7}{2 ( - 2 )} : - 3

\frac{- ( - 5 ) + 7}{2 ( - 2 )}

去除括号:

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

= \frac{5 + 7}{- 2 \cdot 2}

数字相加:

5 + 7 = 12

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

数字相乘:

2 \cdot 2 = 4

= \frac{12}{- 4}

使用分式法则:

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

= - \frac{12}{4}

数字相除:

\frac{12}{4} = 3

= - 3

u = \frac{- ( - 5 ) - 7}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 5 ) - 7}{2 ( - 2 )}

去除括号:

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

= \frac{5 - 7}{- 2 \cdot 2}

数字相减:

5 - 7 = - 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}

二次方程组的解是:

u = - 3, u = \frac{1}{2}

u = cos (θ)

代回

cos (θ) = - 3, cos (θ) = \frac{1}{2}

cos (θ) = - 3, cos (θ) = \frac{1}{2}

cos (θ) = - 3 :

无解

cos (θ) = - 3

- 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

流行的例子

sec^2(x)cos(x)=cos(x),0<x<360

sec^{2} (x) cos (x) = cos (x), 0^{\circ} < x < 36 0^{\circ}

5sin(x)-4=2cos(2x)

5 sin (x) - 4 = 2 cos (2 x)

csc(x)+cot(x)=0

csc (x) + cot (x) = 0

cos^2(θ)-sin^2(θ)=0

cos^{2} (θ) - sin^{2} (θ) = 0

7=4sin(pi/2 (x-2))+5

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