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受欢迎的三角函数 >

sec^2(θ)tan(θ)=2tan(θ),0<= θ<= 360

解答

sec^{2} (θ) tan (θ) = 2 tan (θ), 0^{\circ} \leq θ \leq 36 0^{\circ}

解答

θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}, θ = 13 5^{\circ}, θ = 22 5^{\circ}, θ = 4 5^{\circ}, θ = 31 5^{\circ}

+1

弧度

θ = 0, θ = π, θ = 2 π, θ = \frac{3 π}{4}, θ = \frac{5 π}{4}, θ = \frac{π}{4}, θ = \frac{7 π}{4}

求解步骤

sec^{2} (θ) tan (θ) = 2 tan (θ), 0^{\circ} \leq θ \leq 36 0^{\circ}

两边减去

2 tan (θ)

sec^{2} (θ) tan (θ) - 2 tan (θ) = 0

分解

sec^{2} (θ) tan (θ) - 2 tan (θ) : tan (θ) (sec (θ) + 2) (sec (θ) - 2)

sec^{2} (θ) tan (θ) - 2 tan (θ)

因式分解出通项

tan (θ)

= tan (θ) (sec^{2} (θ) - 2)

分解

sec^{2} (θ) - 2 : (sec (θ) + 2) (sec (θ) - 2)

sec^{2} (θ) - 2

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

= sec^{2} (θ) - (2)^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

sec^{2} (θ) - (2)^{2} = (sec (θ) + 2) (sec (θ) - 2)

= (sec (θ) + 2) (sec (θ) - 2)

= tan (θ) (sec (θ) + 2) (sec (θ) - 2)

tan (θ) (sec (θ) + 2) (sec (θ) - 2) = 0

分别求解每个部分

tan (θ) = 0 or sec (θ) + 2 = 0 or sec (θ) - 2 = 0

tan (θ) = 0, 0 \leq θ \leq 36 0^{\circ} : θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}

tan (θ) = 0, 0 \leq θ \leq 36 0^{\circ}

tan (θ) = 0

的通解

tan (x)

周期表（周期为

18 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = 0 + 18 0^{\circ} n

θ = 0 + 18 0^{\circ} n

解

θ = 0 + 18 0^{\circ} n : θ = 18 0^{\circ} n

θ = 0 + 18 0^{\circ} n

0 + 18 0^{\circ} n = 18 0^{\circ} n

θ = 18 0^{\circ} n

θ = 18 0^{\circ} n

在

0 \leq θ \leq 36 0^{\circ}

范围内的解

θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}

sec (θ) + 2 = 0, 0 \leq θ \leq 36 0^{\circ} : θ = 13 5^{\circ}, θ = 22 5^{\circ}

sec (θ) + 2 = 0, 0 \leq θ \leq 36 0^{\circ}

将

2

到右边

sec (θ) + 2 = 0

两边减去

2

sec (θ) + 2 - 2 = 0 - 2

化简

sec (θ) = - 2

sec (θ) = - 2

sec (θ) = - 2

的通解

sec (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = 13 5^{\circ} + 36 0^{\circ} n, θ = 22 5^{\circ} + 36 0^{\circ} n

θ = 13 5^{\circ} + 36 0^{\circ} n, θ = 22 5^{\circ} + 36 0^{\circ} n

在

0 \leq θ \leq 36 0^{\circ}

范围内的解

θ = 13 5^{\circ}, θ = 22 5^{\circ}

sec (θ) - 2 = 0, 0 \leq θ \leq 36 0^{\circ} : θ = 4 5^{\circ}, θ = 31 5^{\circ}

sec (θ) - 2 = 0, 0 \leq θ \leq 36 0^{\circ}

将

2

到右边

sec (θ) - 2 = 0

两边加上

2

sec (θ) - 2 + 2 = 0 + 2

化简

sec (θ) = 2

sec (θ) = 2

sec (θ) = 2

的通解

sec (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

在

0 \leq θ \leq 36 0^{\circ}

范围内的解

θ = 4 5^{\circ}, θ = 31 5^{\circ}

合并所有解

θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}, θ = 13 5^{\circ}, θ = 22 5^{\circ}, θ = 4 5^{\circ}, θ = 31 5^{\circ}

作图

流行的例子

arccos (x) = 18 0^{\circ}

sin(a)=(29.7)/(54.5)

sin (a) = \frac{29.7}{54.5}

1.0003sin(30)=1.46sin(x)

1.0003 sin (3 0^{\circ}) = 1.46 sin (x)

cos(x)=-5/(sqrt(34))

cos (x) = - \frac{5}{34}

cos(x)=cos(x-c)

cos (x) = cos (x - c)