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受欢迎的三角函数 >

4sin^2(θ)+2cos(θ)=3

解答

4 sin^{2} (θ) + 2 cos (θ) = 3

解答

θ = 1.88495 \dots + 2 πn, θ = - 1.88495 \dots + 2 πn, θ = 0.62831 \dots + 2 πn, θ = 2 π - 0.62831 \dots + 2 πn

+1

度数

θ = 10 8^{\circ} + 36 0^{\circ} n, θ = - 10 8^{\circ} + 36 0^{\circ} n, θ = 3 6^{\circ} + 36 0^{\circ} n, θ = 32 4^{\circ} + 36 0^{\circ} n

求解步骤

4 sin^{2} (θ) + 2 cos (θ) = 3

两边减去

3

4 sin^{2} (θ) + 2 cos (θ) - 3 = 0

使用三角恒等式改写

- 3 + 2 cos (θ) + 4 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) + 4 (1 - cos^{2} (θ))

化简

- 3 + 2 cos (θ) + 4 (1 - cos^{2} (θ)) : 2 cos (θ) - 4 cos^{2} (θ) + 1

- 3 + 2 cos (θ) + 4 (1 - cos^{2} (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)

= - 3 + 2 cos (θ) + 4 - 4 cos^{2} (θ)

化简

- 3 + 2 cos (θ) + 4 - 4 cos^{2} (θ) : 2 cos (θ) - 4 cos^{2} (θ) + 1

- 3 + 2 cos (θ) + 4 - 4 cos^{2} (θ)

对同类项分组

= 2 cos (θ) - 4 cos^{2} (θ) - 3 + 4

数字相加/相减:

- 3 + 4 = 1

= 2 cos (θ) - 4 cos^{2} (θ) + 1

= 2 cos (θ) - 4 cos^{2} (θ) + 1

= 2 cos (θ) - 4 cos^{2} (θ) + 1

1 + 2 cos (θ) - 4 cos^{2} (θ) = 0

用替代法求解

1 + 2 cos (θ) - 4 cos^{2} (θ) = 0

令

: cos (θ) = u

1 + 2 u - 4 u^{2} = 0

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

1 + 2 u - 4 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 4, b = 2, c = 1

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

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

2^{2} - 4 (- 4) \cdot 1 = 25

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

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 4 \cdot 1

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 2^{2} + 16

2^{2} = 4

= 4 + 16

数字相加:

4 + 16 = 20

= 20

20

质因数分解:

2^{2} \cdot 5

20

20

除以

220 = 10 \cdot 2

= 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

使用根式运算法则:

n ab = n a n b

= 5 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 25

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

将解分隔开

u_{1} = \frac{- 2 + 2 5}{2 ( - 4 )}, u_{2} = \frac{- 2 - 2 5}{2 ( - 4 )}

u = \frac{- 2 + 2 5}{2 ( - 4 )} : - \frac{- 1 + 5}{4}

\frac{- 2 + 2 5}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 2 + 2 5}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 2 + 2 5}{- 8}

使用分式法则:

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

= - \frac{- 2 + 2 5}{8}

消掉

\frac{- 2 + 2 5}{8} : \frac{5 - 1}{4}

\frac{- 2 + 2 5}{8}

分解

- 2 + 25 : 2 (- 1 + 5)

- 2 + 25

改写为

= - 2 \cdot 1 + 25

因式分解出通项

2

= 2 (- 1 + 5)

= \frac{2 ( - 1 + 5 )}{8}

约分:

2

= \frac{- 1 + 5}{4}

= - \frac{5 - 1}{4}

= - \frac{- 1 + 5}{4}

u = \frac{- 2 - 2 5}{2 ( - 4 )} : \frac{1 + 5}{4}

\frac{- 2 - 2 5}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 2 - 2 5}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 2 - 2 5}{- 8}

使用分式法则:

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

- 2 - 25 = - (2 + 25)

= \frac{2 + 2 5}{8}

分解

2 + 25 : 2 (1 + 5)

2 + 25

改写为

= 2 \cdot 1 + 25

因式分解出通项

2

= 2 (1 + 5)

= \frac{2 ( 1 + 5 )}{8}

约分:

2

= \frac{1 + 5}{4}

二次方程组的解是:

u = - \frac{- 1 + 5}{4}, u = \frac{1 + 5}{4}

u = cos (θ)

代回

cos (θ) = - \frac{- 1 + 5}{4}, cos (θ) = \frac{1 + 5}{4}

cos (θ) = - \frac{- 1 + 5}{4}, cos (θ) = \frac{1 + 5}{4}

cos (θ) = - \frac{- 1 + 5}{4} : θ = arccos (- \frac{- 1 + 5}{4}) + 2 πn, θ = - arccos (- \frac{- 1 + 5}{4}) + 2 πn

cos (θ) = - \frac{- 1 + 5}{4}

使用反三角函数性质

cos (θ) = - \frac{- 1 + 5}{4}

cos (θ) = - \frac{- 1 + 5}{4}

的通解

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

θ = arccos (- \frac{- 1 + 5}{4}) + 2 πn, θ = - arccos (- \frac{- 1 + 5}{4}) + 2 πn

θ = arccos (- \frac{- 1 + 5}{4}) + 2 πn, θ = - arccos (- \frac{- 1 + 5}{4}) + 2 πn

cos (θ) = \frac{1 + 5}{4} : θ = arccos (\frac{1 + 5}{4}) + 2 πn, θ = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

cos (θ) = \frac{1 + 5}{4}

使用反三角函数性质

cos (θ) = \frac{1 + 5}{4}

cos (θ) = \frac{1 + 5}{4}

的通解

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

θ = arccos (\frac{1 + 5}{4}) + 2 πn, θ = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

θ = arccos (\frac{1 + 5}{4}) + 2 πn, θ = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

合并所有解

θ = arccos (- \frac{- 1 + 5}{4}) + 2 πn, θ = - arccos (- \frac{- 1 + 5}{4}) + 2 πn, θ = arccos (\frac{1 + 5}{4}) + 2 πn, θ = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

以小数形式表示解

θ = 1.88495 \dots + 2 πn, θ = - 1.88495 \dots + 2 πn, θ = 0.62831 \dots + 2 πn, θ = 2 π - 0.62831 \dots + 2 πn

作图

流行的例子

1715/70 =tan(45+θ/2)

\frac{1715}{70} = tan (4 5^{\circ} + \frac{θ}{2})

cos(3x)=sin(x-30)

cos (3 x) = sin (x - 3 0^{\circ})

885cos(θ)-70=cos(250)

885 cos (θ) - 70 = cos (25 0^{\circ})

tan(x)+5=0,x<= 0<2pi

tan (x) + 5 = 0, x \leq 0 < 2 π

2sqrt(3)cos(θ-pi/3)=3,0<= θ<= 2pi

23 cos (θ - \frac{π}{3}) = 3, 0 \leq θ \leq 2 π