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受欢迎的三角函数 >

sin^2(2θ)-5sin(2θ)-1=0

解答

sin^{2} (2 θ) - 5 sin (2 θ) - 1 = 0

解答

θ = \frac{- 0.19379 \dots}{2} + πn, θ = \frac{π}{2} + \frac{0.19379 \dots}{2} + πn

+1

度数

θ = - 5.55176 \dots^{\circ} + 18 0^{\circ} n, θ = 95.55176 \dots^{\circ} + 18 0^{\circ} n

求解步骤

sin^{2} (2 θ) - 5 sin (2 θ) - 1 = 0

用替代法求解

sin^{2} (2 θ) - 5 sin (2 θ) - 1 = 0

令

: sin (2 θ) = u

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

u^{2} - 5 u - 1 = 0 : u = \frac{5 + 29}{2}, u = \frac{5 - 29}{2}

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 5)^{2} - 4 \cdot 1 \cdot (- 1) = 29

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 5^{2} + 4

5^{2} = 25

= 25 + 4

数字相加:

25 + 4 = 29

= 29

u_{1, 2} = \frac{- ( - 5 ) \pm 29}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 29}{2 \cdot 1}, u_{2} = \frac{- ( - 5 ) - 29}{2 \cdot 1}

u = \frac{- ( - 5 ) + 29}{2 \cdot 1} : \frac{5 + 29}{2}

\frac{- ( - 5 ) + 29}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{5 + 29}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{5 + 29}{2}

u = \frac{- ( - 5 ) - 29}{2 \cdot 1} : \frac{5 - 29}{2}

\frac{- ( - 5 ) - 29}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{5 - 29}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{5 - 29}{2}

二次方程组的解是:

u = \frac{5 + 29}{2}, u = \frac{5 - 29}{2}

u = sin (2 θ)

代回

sin (2 θ) = \frac{5 + 29}{2}, sin (2 θ) = \frac{5 - 29}{2}

sin (2 θ) = \frac{5 + 29}{2}, sin (2 θ) = \frac{5 - 29}{2}

sin (2 θ) = \frac{5 + 29}{2} :

无解

sin (2 θ) = \frac{5 + 29}{2}

- 1 \leq sin (x) \leq 1

无解

sin (2 θ) = \frac{5 - 29}{2} : θ = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + πn, θ = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + πn

sin (2 θ) = \frac{5 - 29}{2}

使用反三角函数性质

sin (2 θ) = \frac{5 - 29}{2}

sin (2 θ) = \frac{5 - 29}{2}

的通解

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

2 θ = arcsin (\frac{5 - 29}{2}) + 2 πn, 2 θ = π + arcsin (- \frac{5 - 29}{2}) + 2 πn

2 θ = arcsin (\frac{5 - 29}{2}) + 2 πn, 2 θ = π + arcsin (- \frac{5 - 29}{2}) + 2 πn

解

2 θ = arcsin (\frac{5 - 29}{2}) + 2 πn : θ = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + πn

2 θ = arcsin (\frac{5 - 29}{2}) + 2 πn

两边除以

2

2 θ = arcsin (\frac{5 - 29}{2}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + \frac{2 πn}{2}

化简

θ = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + πn

θ = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + πn

解

2 θ = π + arcsin (- \frac{5 - 29}{2}) + 2 πn : θ = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + πn

2 θ = π + arcsin (- \frac{5 - 29}{2}) + 2 πn

- \frac{5 - 29}{2} = - \frac{- ( 29 - 5 )}{2} = \frac{29 - 5}{2}

2 θ = π + arcsin (\frac{29 - 5}{2}) + 2 πn

两边除以

2

2 θ = π + arcsin (\frac{29 - 5}{2}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + \frac{2 πn}{2}

化简

θ = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + πn

θ = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + πn

θ = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + πn, θ = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + πn

合并所有解

θ = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + πn, θ = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + πn

以小数形式表示解

θ = \frac{- 0.19379 \dots}{2} + πn, θ = \frac{π}{2} + \frac{0.19379 \dots}{2} + πn

作图

流行的例子

-pi/(30)sin(pi/(90)(x+20))=0

- \frac{π}{30} sin (\frac{π}{90} (x + 20)) = 0

sin(2x)=-1/(sqrt(2))

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

cos(x)=-(22)/(7.28*sqrt(20))

cos (x) = - \frac{22}{7.28 \cdot 20}

sec^2(x)=4tan(x)

sec^{2} (x) = 4 tan (x)

solvefor x,1+sin(2x-30)=0

so l v e f or x, 1 + sin (2 x - 3 0^{\circ}) = 0