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受欢迎的三角函数 >

1-2tan^2(θ)=-tan^2(θ)

解答

1 - 2 tan^{2} (θ) = - tan^{2} (θ)

解答

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

+1

度数

θ = 4 5^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

1 - 2 tan^{2} (θ) = - tan^{2} (θ)

用替代法求解

1 - 2 tan^{2} (θ) = - tan^{2} (θ)

令

: tan (θ) = u

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

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

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

将

1

到右边

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

两边减去

1

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

化简

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

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

将

u^{2}

para o lado esquerdo

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

两边加上

u^{2}

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

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 (θ)

代回

tan (θ) = 1, tan (θ) = - 1

tan (θ) = 1, tan (θ) = - 1

tan (θ) = 1 : θ = \frac{π}{4} + πn

tan (θ) = 1

tan (θ) = 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

tan (θ) = - 1 : θ = \frac{3 π}{4} + πn

tan (θ) = - 1

tan (θ) = - 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

合并所有解

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

作图

流行的例子

cos (a) = \frac{1}{3}

2cos(2θ)=sqrt(3)

2 cos (2 θ) = 3

2sin^2(θ)-5sin(θ)+2=0

2 sin^{2} (θ) - 5 sin (θ) + 2 = 0

tan (2 x) + 1 = 0

sin(x)-sqrt(2)=-sin(x)

sin (x) - 2 = - sin (x)