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受欢迎的三角函数 >

2sin(θ)-tan^2(θ)=0

解答

2 sin (θ) - tan^{2} (θ) = 0

解答

θ = 2 πn, θ = π + 2 πn, θ = 0.89590 \dots + 2 πn, θ = π - 0.89590 \dots + 2 πn

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 51.33171 \dots^{\circ} + 36 0^{\circ} n, θ = 128.66828 \dots^{\circ} + 36 0^{\circ} n

求解步骤

2 sin (θ) - tan^{2} (θ) = 0

用 sin, cos 表示

- tan^{2} (θ) + 2 sin (θ)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - (\frac{sin ( θ )}{cos ( θ )})^{2} + 2 sin (θ)

化简

- (\frac{sin ( θ )}{cos ( θ )})^{2} + 2 sin (θ) : \frac{- sin ^{2} ( θ ) + 2 cos ^{2} ( θ ) sin ( θ )}{cos ^{2} ( θ )}

- (\frac{sin ( θ )}{cos ( θ )})^{2} + 2 sin (θ)

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= - \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )} + 2 sin (θ)

将项转换为分式:

2 sin (θ) = \frac{2 sin ( θ ) cos ^{2} ( θ )}{cos ^{2} ( θ )}

= - \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )} + \frac{2 sin ( θ ) cos ^{2} ( θ )}{cos ^{2} ( θ )}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- sin ^{2} ( θ ) + 2 sin ( θ ) cos ^{2} ( θ )}{cos ^{2} ( θ )}

= \frac{- sin ^{2} ( θ ) + 2 cos ^{2} ( θ ) sin ( θ )}{cos ^{2} ( θ )}

\frac{- sin ^{2} ( θ ) + 2 cos ^{2} ( θ ) sin ( θ )}{cos ^{2} ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

分解

- sin^{2} (θ) + 2 cos^{2} (θ) sin (θ) : sin (θ) (- sin (θ) + 2 cos^{2} (θ))

- sin^{2} (θ) + 2 cos^{2} (θ) sin (θ)

使用指数法则:

a^{b + c} = a^{b} a^{c}

sin^{2} (θ) = sin (θ) sin (θ)

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

因式分解出通项

sin (θ)

= sin (θ) (- sin (θ) + 2 cos^{2} (θ))

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

分别求解每个部分

sin (θ) = 0 or - sin (θ) + 2 cos^{2} (θ) = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

sin (θ) = 0

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 = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

- sin (θ) + 2 cos^{2} (θ) = 0 : θ = arcsin (\frac{17 - 1}{4}) + 2 πn, θ = π - arcsin (\frac{17 - 1}{4}) + 2 πn

- sin (θ) + 2 cos^{2} (θ) = 0

使用三角恒等式改写

- sin (θ) + 2 cos^{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)

= - sin (θ) + 2 (1 - sin^{2} (θ))

- sin (θ) + (1 - sin^{2} (θ)) \cdot 2 = 0

用替代法求解

- sin (θ) + (1 - sin^{2} (θ)) \cdot 2 = 0

令

: sin (θ) = u

- u + (1 - u^{2}) \cdot 2 = 0

- u + (1 - u^{2}) \cdot 2 = 0 : u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

- u + (1 - u^{2}) \cdot 2 = 0

展开

- u + (1 - u^{2}) \cdot 2 : - u + 2 - 2 u^{2}

- u + (1 - u^{2}) \cdot 2

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

乘开

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

= 2 \cdot 1 - 2 u^{2}

数字相乘:

2 \cdot 1 = 2

= 2 - 2 u^{2}

= - u + 2 - 2 u^{2}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 1)^{2} - 4 (- 2) \cdot 2 = 17

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

使用法则

- (- a) = a

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

(- 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 2 \cdot 2 = 16

4 \cdot 2 \cdot 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 16

= 1 + 16

数字相加:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 17}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - 17}{2 ( - 2 )}

u = \frac{- ( - 1 ) + 17}{2 ( - 2 )} : - \frac{1 + 17}{4}

\frac{- ( - 1 ) + 17}{2 ( - 2 )}

去除括号:

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

= \frac{1 + 17}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{1 + 17}{- 4}

使用分式法则:

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

= - \frac{1 + 17}{4}

u = \frac{- ( - 1 ) - 17}{2 ( - 2 )} : \frac{17 - 1}{4}

\frac{- ( - 1 ) - 17}{2 ( - 2 )}

去除括号:

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

= \frac{1 - 17}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{1 - 17}{- 4}

使用分式法则:

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

1 - 17 = - (17 - 1)

= \frac{17 - 1}{4}

二次方程组的解是:

u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

u = sin (θ)

代回

sin (θ) = - \frac{1 + 17}{4}, sin (θ) = \frac{17 - 1}{4}

sin (θ) = - \frac{1 + 17}{4}, sin (θ) = \frac{17 - 1}{4}

sin (θ) = - \frac{1 + 17}{4} :

无解

sin (θ) = - \frac{1 + 17}{4}

- 1 \leq sin (x) \leq 1

无解

sin (θ) = \frac{17 - 1}{4} : θ = arcsin (\frac{17 - 1}{4}) + 2 πn, θ = π - arcsin (\frac{17 - 1}{4}) + 2 πn

sin (θ) = \frac{17 - 1}{4}

使用反三角函数性质

sin (θ) = \frac{17 - 1}{4}

sin (θ) = \frac{17 - 1}{4}

的通解

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

θ = arcsin (\frac{17 - 1}{4}) + 2 πn, θ = π - arcsin (\frac{17 - 1}{4}) + 2 πn

θ = arcsin (\frac{17 - 1}{4}) + 2 πn, θ = π - arcsin (\frac{17 - 1}{4}) + 2 πn

合并所有解

θ = arcsin (\frac{17 - 1}{4}) + 2 πn, θ = π - arcsin (\frac{17 - 1}{4}) + 2 πn

合并所有解

θ = 2 πn, θ = π + 2 πn, θ = arcsin (\frac{17 - 1}{4}) + 2 πn, θ = π - arcsin (\frac{17 - 1}{4}) + 2 πn

以小数形式表示解

θ = 2 πn, θ = π + 2 πn, θ = 0.89590 \dots + 2 πn, θ = π - 0.89590 \dots + 2 πn

作图

流行的例子

sin(θ)-sqrt(3)cos(θ)=1

sin (θ) - 3 cos (θ) = 1

cos (a) = \frac{1}{2}

cos^{2} (θ) = \frac{3}{4}

tan (θ) = \frac{3}{8}

9cos(x)+5=-cos(x)

9 cos (x) + 5 = - cos (x)