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受欢迎的三角函数 >

5-tan^2(x)=4,0<= x<= 360

解答

5 - tan^{2} (x) = 4, 0^{\circ} \leq x \leq 36 0^{\circ}

解答

x = 4 5^{\circ}, x = 22 5^{\circ}, x = 13 5^{\circ}, x = 31 5^{\circ}

+1

弧度

x = \frac{π}{4}, x = \frac{5 π}{4}, x = \frac{3 π}{4}, x = \frac{7 π}{4}

求解步骤

5 - tan^{2} (x) = 4, 0^{\circ} \leq x \leq 36 0^{\circ}

用替代法求解

5 - tan^{2} (x) = 4

令

: tan (x) = u

5 - u^{2} = 4

5 - u^{2} = 4 : u = 1, u = - 1

5 - u^{2} = 4

将

5

到右边

5 - u^{2} = 4

两边减去

5

5 - u^{2} - 5 = 4 - 5

化简

- u^{2} = - 1

- u^{2} = - 1

两边除以

- 1

- u^{2} = - 1

两边除以

- 1

\frac{- u ^{2}}{- 1} = \frac{- 1}{- 1}

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

u = tan (x)

代回

tan (x) = 1, tan (x) = - 1

tan (x) = 1, tan (x) = - 1

tan (x) = 1, 0 \leq x \leq 36 0^{\circ} : x = 4 5^{\circ}, x = 22 5^{\circ}

tan (x) = 1, 0 \leq x \leq 36 0^{\circ}

tan (x) = 1

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}

x = 4 5^{\circ} + 18 0^{\circ} n

x = 4 5^{\circ} + 18 0^{\circ} n

在

0 \leq x \leq 36 0^{\circ}

范围内的解

x = 4 5^{\circ}, x = 22 5^{\circ}

tan (x) = - 1, 0 \leq x \leq 36 0^{\circ} : x = 13 5^{\circ}, x = 31 5^{\circ}

tan (x) = - 1, 0 \leq x \leq 36 0^{\circ}

tan (x) = - 1

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}

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

在

0 \leq x \leq 36 0^{\circ}

范围内的解

x = 13 5^{\circ}, x = 31 5^{\circ}

合并所有解

x = 4 5^{\circ}, x = 22 5^{\circ}, x = 13 5^{\circ}, x = 31 5^{\circ}

作图

流行的例子

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sin (x - \frac{π}{3}) = \frac{1}{2}

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

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