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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}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

+1

度数

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\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 = \frac{2}{2}, u = 0

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

42 \cdot 0 = 0

42 \cdot 0

使用法则

0 \cdot a = 0

= 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} : \frac{2}{2}

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

使用法则

- (- a) = a

= \frac{1 + 1}{2 2}

数字相加:

1 + 1 = 2

= \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}

u = \frac{- ( - 1 ) - 1}{2 2} : 0

\frac{- ( - 1 ) - 1}{2 2}

使用法则

- (- a) = a

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

数字相减:

1 - 1 = 0

= \frac{0}{2 2}

使用法则

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

= 0

二次方程组的解是:

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

u = cos (θ)

代回

cos (θ) = \frac{2}{2}, cos (θ) = 0

cos (θ) = \frac{2}{2}, cos (θ) = 0

cos (θ) = \frac{2}{2} : θ = \frac{π}{4} + 2 πn, θ = \frac{7 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{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

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

合并所有解

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

作图

流行的例子

cos^2(x)+cos(x)-2=0

cos^{2} (x) + cos (x) - 2 = 0

cos(x)=sqrt(1/2)

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

cos(x)=-1,0<= x<= 2pi

cos (x) = - 1, 0 \leq x \leq 2 π

sin(x)tan^2(x)=sin(x)

sin (x) tan^{2} (x) = sin (x)

sin(θ)cos(3θ)+cos(θ)sin(3θ)=0

sin (θ) cos (3 θ) + cos (θ) sin (3 θ) = 0