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受欢迎的三角函数 >

6tan^2(θ)-5tan(θ)+1=0

解答

6 tan^{2} (θ) - 5 tan (θ) + 1 = 0

解答

θ = 0.46364 \dots + πn, θ = 0.32175 \dots + πn

+1

度数

θ = 26.56505 \dots^{\circ} + 18 0^{\circ} n, θ = 18.43494 \dots^{\circ} + 18 0^{\circ} n

求解步骤

6 tan^{2} (θ) - 5 tan (θ) + 1 = 0

用替代法求解

6 tan^{2} (θ) - 5 tan (θ) + 1 = 0

令

: tan (θ) = u

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

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

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 \cdot 6 \cdot 1}{2 \cdot 6}

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

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

(- 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 \cdot 6}

将解分隔开

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

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

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

使用法则

- (- 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}

约分:

6

= \frac{1}{2}

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

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

使用法则

- (- 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}

约分:

4

= \frac{1}{3}

二次方程组的解是:

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

u = tan (θ)

代回

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

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

tan (θ) = \frac{1}{2} : θ = arctan (\frac{1}{2}) + πn

tan (θ) = \frac{1}{2}

使用反三角函数性质

tan (θ) = \frac{1}{2}

tan (θ) = \frac{1}{2}

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

θ = arctan (\frac{1}{2}) + πn

θ = arctan (\frac{1}{2}) + πn

tan (θ) = \frac{1}{3} : θ = arctan (\frac{1}{3}) + πn

tan (θ) = \frac{1}{3}

使用反三角函数性质

tan (θ) = \frac{1}{3}

tan (θ) = \frac{1}{3}

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

θ = arctan (\frac{1}{3}) + πn

θ = arctan (\frac{1}{3}) + πn

合并所有解

θ = arctan (\frac{1}{2}) + πn, θ = arctan (\frac{1}{3}) + πn

以小数形式表示解

θ = 0.46364 \dots + πn, θ = 0.32175 \dots + πn

作图

流行的例子

6 sin (x) - 24 = 0

-2sin(x)-2sin(2x)=0

- 2 sin (x) - 2 sin (2 x) = 0

((sin^2(x)+1))/(1-sin^2(x))-1=2

\frac{( sin ^{2} ( x ) + 1 )}{1 - sin ^{2} ( x )} - 1 = 2

-2sin(θ)+cos(2θ)=1

- 2 sin (θ) + cos (2 θ) = 1

cos(2x+60)=sin(3x)

cos (2 x + 60) = sin (3 x)