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受欢迎的三角函数 >

sec^2(x)-2=tan^2(x),0<= x<= 2pi

解答

sec^{2} (x) - 2 = tan^{2} (x), 0 \leq x \leq 2 π

解答

x \in R 无解

求解步骤

sec^{2} (x) - 2 = tan^{2} (x), 0 \leq x \leq 2 π

两边减去

tan^{2} (x)

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

用 sin, cos 表示

- 2 + sec^{2} (x) - tan^{2} (x)

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

= - 2 + (\frac{1}{cos ( x )})^{2} - tan^{2} (x)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

化简

- 2 + (\frac{1}{cos ( x )})^{2} - (\frac{sin ( x )}{cos ( x )})^{2} : \frac{- 2 cos ^{2} ( x ) + 1 - sin ^{2} ( x )}{cos ^{2} ( x )}

- 2 + (\frac{1}{cos ( x )})^{2} - (\frac{sin ( x )}{cos ( x )})^{2}

(\frac{1}{cos ( x )})^{2} = \frac{1}{cos ^{2} ( x )}

(\frac{1}{cos ( x )})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cos ^{2} ( x )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )})^{2} = \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

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

合并分式

\frac{1}{cos ^{2} ( x )} - \frac{sin ^{2} ( x )}{cos ^{2} ( x )} : \frac{1 - sin ^{2} ( x )}{cos ^{2} ( x )}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

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

将项转换为分式:

2 = \frac{2 cos ^{2} ( x )}{cos ^{2} ( x )}

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

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

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

\frac{1 - sin ^{2} ( x ) - 2 cos ^{2} ( x )}{cos ^{2} ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

1 - sin^{2} (x) - 2 cos^{2} (x) = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

= - cos^{2} (x)

- cos^{2} (x) = 0

两边除以

- 1

- cos^{2} (x) = 0

两边除以

- 1

- cos^{2} (x) = 0

两边除以

- 1

\frac{- cos ^{2} ( x )}{- 1} = \frac{0}{- 1}

化简

cos^{2} (x) = 0

cos^{2} (x) = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

cos (x) = 0

cos (x) = 0

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = \frac{π}{2}, x = \frac{3 π}{2}

因为方程对以下值无定义:

\frac{π}{2}, \frac{3 π}{2}

x \in R 无解

作图

流行的例子

4cos(x)-3cos(x)-1=0

4 cos (x) - 3 cos (x) - 1 = 0

3sin(60-(3x)/4)=5sin((3x)/4-30)

3 sin (6 0^{\circ} - \frac{3 x}{4}) = 5 sin (\frac{3 x}{4} - 3 0^{\circ})

(sin(θ))/(60)=(sin(80))/(280.87)

\frac{sin ( θ )}{60} = \frac{sin ( 8 0 ^{\circ} )}{280.87}

5.63=11cos((2pi)/T 0.137)

5.63 = 11 cos (\frac{2 π}{T} 0.137)

solvefor x=2arctan(z),z

so l v e f or x = 2 arctan (z), z