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受欢迎的三角函数 >

3cos(2θ)-4sin(θ)+2=sin(θ)+6

解答

3 cos (2 θ) - 4 sin (θ) + 2 = sin (θ) + 6

解答

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = - 0.33983 \dots + 2 πn, θ = π + 0.33983 \dots + 2 πn

+1

度数

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n, θ = - 19.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 199.47122 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 cos (2 θ) - 4 sin (θ) + 2 = sin (θ) + 6

两边减去

sin (θ) + 6

3 cos (2 θ) - 5 sin (θ) - 4 = 0

使用三角恒等式改写

- 4 + 3 cos (2 θ) - 5 sin (θ)

使用倍角公式:

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

= - 4 + 3 (1 - 2 sin^{2} (θ)) - 5 sin (θ)

化简

- 4 + 3 (1 - 2 sin^{2} (θ)) - 5 sin (θ) : - 6 sin^{2} (θ) - 5 sin (θ) - 1

- 4 + 3 (1 - 2 sin^{2} (θ)) - 5 sin (θ)

乘开

3 (1 - 2 sin^{2} (θ)) : 3 - 6 sin^{2} (θ)

3 (1 - 2 sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = 2 sin^{2} (θ)

= 3 \cdot 1 - 3 \cdot 2 sin^{2} (θ)

化简

3 \cdot 1 - 3 \cdot 2 sin^{2} (θ) : 3 - 6 sin^{2} (θ)

3 \cdot 1 - 3 \cdot 2 sin^{2} (θ)

数字相乘:

3 \cdot 1 = 3

= 3 - 3 \cdot 2 sin^{2} (θ)

数字相乘:

3 \cdot 2 = 6

= 3 - 6 sin^{2} (θ)

= 3 - 6 sin^{2} (θ)

= - 4 + 3 - 6 sin^{2} (θ) - 5 sin (θ)

数字相加/相减:

- 4 + 3 = - 1

= - 6 sin^{2} (θ) - 5 sin (θ) - 1

= - 6 sin^{2} (θ) - 5 sin (θ) - 1

- 1 - 5 sin (θ) - 6 sin^{2} (θ) = 0

用替代法求解

- 1 - 5 sin (θ) - 6 sin^{2} (θ) = 0

令

: sin (θ) = u

- 1 - 5 u - 6 u^{2} = 0

- 1 - 5 u - 6 u^{2} = 0 : u = - \frac{1}{2}, u = - \frac{1}{3}

- 1 - 5 u - 6 u^{2} = 0

改写成标准形式

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

- 6 u^{2} - 5 u - 1 = 0

使用求根公式求解

- 6 u^{2} - 5 u - 1 = 0

二次方程求根公式:

若

a = - 6, b = - 5, c = - 1

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 6 ) ( - 1 )}{2 ( - 6 )}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 6 ) ( - 1 )}{2 ( - 6 )}

(- 5)^{2} - 4 (- 6) (- 1) = 1

(- 5)^{2} - 4 (- 6) (- 1)

使用法则

- (- a) = a

= (- 5)^{2} - 4 \cdot 6 \cdot 1

使用指数法则:

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

若

n

是偶数

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

= 5^{2} - 4 \cdot 6 \cdot 1

数字相乘:

4 \cdot 6 \cdot 1 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

数字相减:

25 - 24 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 ( - 6 )}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 1}{2 ( - 6 )}, u_{2} = \frac{- ( - 5 ) - 1}{2 ( - 6 )}

u = \frac{- ( - 5 ) + 1}{2 ( - 6 )} : - \frac{1}{2}

\frac{- ( - 5 ) + 1}{2 ( - 6 )}

去除括号:

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

= \frac{5 + 1}{- 2 \cdot 6}

数字相加:

5 + 1 = 6

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

数字相乘:

2 \cdot 6 = 12

= \frac{6}{- 12}

使用分式法则:

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

= - \frac{6}{12}

约分:

6

= - \frac{1}{2}

u = \frac{- ( - 5 ) - 1}{2 ( - 6 )} : - \frac{1}{3}

\frac{- ( - 5 ) - 1}{2 ( - 6 )}

去除括号:

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

= \frac{5 - 1}{- 2 \cdot 6}

数字相减:

5 - 1 = 4

= \frac{4}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{4}{- 12}

使用分式法则:

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

= - \frac{4}{12}

约分:

4

= - \frac{1}{3}

二次方程组的解是:

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

u = sin (θ)

代回

sin (θ) = - \frac{1}{2}, sin (θ) = - \frac{1}{3}

sin (θ) = - \frac{1}{2}, sin (θ) = - \frac{1}{3}

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2}

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

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

合并所有解

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = arcsin (- \frac{1}{3}) + 2 πn, θ = π + arcsin (\frac{1}{3}) + 2 πn

以小数形式表示解

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = - 0.33983 \dots + 2 πn, θ = π + 0.33983 \dots + 2 πn

作图

流行的例子

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2 cos (x) sin (x) + cos (x) = 0

6cos(x)-6sin(x)=3sqrt(6)

6 cos (x) - 6 sin (x) = 36

cos^2(θ)+cos^4(θ)=1

cos^{2} (θ) + cos^{4} (θ) = 1

cot(a)=-1/(sqrt(3))

cot (a) = - \frac{1}{3}

2cos(θ)-2sec(θ)-3=0

2 cos (θ) - 2 sec (θ) - 3 = 0