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受欢迎的三角函数 >

-2cos(2θ)+12=4sin(θ)+9

解答

- 2 cos (2 θ) + 12 = 4 sin (θ) + 9

解答

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

+1

度数

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

- 2 cos (2 θ) + 12 = 4 sin (θ) + 9

两边减去

4 sin (θ) + 9

- 2 cos (2 θ) - 4 sin (θ) + 3 = 0

使用三角恒等式改写

3 - 2 cos (2 θ) - 4 sin (θ)

使用倍角公式:

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

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

化简

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

3 - 2 (1 - 2 sin^{2} (θ)) - 4 sin (θ)

乘开

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

- 2 (1 - 2 sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 2, b = 1, c = 2 sin^{2} (θ)

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

使用加减运算法则

- (- a) = a

= - 2 \cdot 1 + 2 \cdot 2 sin^{2} (θ)

化简

- 2 \cdot 1 + 2 \cdot 2 sin^{2} (θ) : - 2 + 4 sin^{2} (θ)

- 2 \cdot 1 + 2 \cdot 2 sin^{2} (θ)

数字相乘:

2 \cdot 1 = 2

= - 2 + 2 \cdot 2 sin^{2} (θ)

数字相乘:

2 \cdot 2 = 4

= - 2 + 4 sin^{2} (θ)

= - 2 + 4 sin^{2} (θ)

= 3 - 2 + 4 sin^{2} (θ) - 4 sin (θ)

数字相减:

3 - 2 = 1

= 4 sin^{2} (θ) - 4 sin (θ) + 1

= 4 sin^{2} (θ) - 4 sin (θ) + 1

1 - 4 sin (θ) + 4 sin^{2} (θ) = 0

用替代法求解

1 - 4 sin (θ) + 4 sin^{2} (θ) = 0

令

: sin (θ) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 4)^{2} - 4 \cdot 4 \cdot 1 = 0

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

使用指数法则:

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

若

n

是偶数

(- 4)^{2} = 4^{2}

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

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

数字相减:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 4}

u = \frac{- ( - 4 )}{2 \cdot 4}

\frac{- ( - 4 )}{2 \cdot 4} = \frac{1}{2}

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 4 = 8

= \frac{4}{8}

约分:

4

= \frac{1}{2}

u = \frac{1}{2}

二次方程组的解是:

u = \frac{1}{2}

u = sin (θ)

代回

sin (θ) = \frac{1}{2}

sin (θ) = \frac{1}{2}

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 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{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

合并所有解

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

作图

流行的例子

sec^2(x)+sec(x)-2=0

sec^{2} (x) + sec (x) - 2 = 0

(cot(x)-1)(2sin(x)+1)=0

(cot (x) - 1) (2 sin (x) + 1) = 0

solvefor y,x=tan(y)

so l v e f or y, x = tan (y)

tan (x) = - \frac{12}{5}

2 cos (x) = sin (x)