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受欢迎的三角函数 >

4sec(θ)=-5-tan^2(θ)

解答

4 sec (θ) = - 5 - tan^{2} (θ)

解答

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

+1

度数

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

求解步骤

4 sec (θ) = - 5 - tan^{2} (θ)

两边减去

- 5 - tan^{2} (θ)

4 sec (θ) + 5 + tan^{2} (θ) = 0

使用三角恒等式改写

5 + tan^{2} (θ) + 4 sec (θ)

使用毕达哥拉斯恒等式:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

= 5 + sec^{2} (θ) - 1 + 4 sec (θ)

化简

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

5 + sec^{2} (θ) - 1 + 4 sec (θ)

对同类项分组

= sec^{2} (θ) + 4 sec (θ) + 5 - 1

数字相加/相减:

5 - 1 = 4

= sec^{2} (θ) + 4 sec (θ) + 4

= sec^{2} (θ) + 4 sec (θ) + 4

4 + sec^{2} (θ) + 4 sec (θ) = 0

用替代法求解

4 + sec^{2} (θ) + 4 sec (θ) = 0

令

: sec (θ) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 1, b = 4, c = 4

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

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

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

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

数字相乘:

4 \cdot 1 \cdot 4 = 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 1}

u = \frac{- 4}{2 \cdot 1}

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

\frac{- 4}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 4}{2}

使用分式法则:

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

= - \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= - 2

u = - 2

二次方程组的解是:

u = - 2

u = sec (θ)

代回

sec (θ) = - 2

sec (θ) = - 2

sec (θ) = - 2 : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

sec (θ) = - 2

sec (θ) = - 2

的通解

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

合并所有解

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

作图

流行的例子

cos(2x)=1-3sin(x)

cos (2 x) = 1 - 3 sin (x)

3sin(2t)+4=1,-pi/2 <= t<= pi/2

3 sin (2 t) + 4 = 1, - \frac{π}{2} \leq t \leq \frac{π}{2}

2cos(x)+sin(x)=1

2 cos (x) + sin (x) = 1

5sin(x)=5sin(2x)

5 sin (x) = 5 sin (2 x)

- 0.6 = sin (2 θ)