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受欢迎的三角函数 >

sqrt(2)cos^2(θ)+cos(θ)=0

解答

2 cos^{2} (θ) + cos (θ) = 0

解答

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

+1

度数

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 36 0^{\circ} n, θ = 22 5^{\circ} + 36 0^{\circ} n

求解步骤

2 cos^{2} (θ) + cos (θ) = 0

用替代法求解

2 cos^{2} (θ) + cos (θ) = 0

令

: cos (θ) = u

2 u^{2} + u = 0

2 u^{2} + u = 0 : u = 0, u = - \frac{2}{2}

2 u^{2} + u = 0

使用求根公式求解

2 u^{2} + u = 0

二次方程求根公式:

若

a = 2, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 2 \cdot 0}{2 2}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 2 \cdot 0}{2 2}

1^{2} - 42 \cdot 0 = 1

1^{2} - 42 \cdot 0

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 0 \cdot 2

使用法则

0 \cdot a = 0

= 1 - 0

数字相减:

1 - 0 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- 1 \pm 1}{2 2}

将解分隔开

u_{1} = \frac{- 1 + 1}{2 2}, u_{2} = \frac{- 1 - 1}{2 2}

u = \frac{- 1 + 1}{2 2} : 0

\frac{- 1 + 1}{2 2}

数字相加/相减:

- 1 + 1 = 0

= \frac{0}{2 2}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

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

\frac{- 1 - 1}{2 2}

数字相减:

- 1 - 1 = - 2

= \frac{- 2}{2 2}

使用分式法则:

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

= - \frac{2}{2 2}

数字相除:

\frac{2}{2} = 1

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

二次方程组的解是:

u = 0, u = - \frac{2}{2}

u = cos (θ)

代回

cos (θ) = 0, cos (θ) = - \frac{2}{2}

cos (θ) = 0, cos (θ) = - \frac{2}{2}

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

cos (θ) = 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}

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

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

cos (θ) = - \frac{2}{2} : θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

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}

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

合并所有解

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

作图

流行的例子

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9 sec (x) + 9 tan (x) = 9

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