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受欢迎的三角函数 >

4cos^2(θ)=1,0<= θ<2pi

解答

4 cos^{2} (θ) = 1, 0 \leq θ < 2 π

解答

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

+1

度数

θ = 6 0^{\circ}, θ = 30 0^{\circ}, θ = 12 0^{\circ}, θ = 24 0^{\circ}

求解步骤

4 cos^{2} (θ) = 1, 0 \leq θ < 2 π

用替代法求解

4 cos^{2} (θ) = 1

令

: cos (θ) = u

4 u^{2} = 1

4 u^{2} = 1 : u = \frac{1}{2}, u = - \frac{1}{2}

4 u^{2} = 1

两边除以

4

4 u^{2} = 1

两边除以

4

\frac{4 u ^{2}}{4} = \frac{1}{4}

化简

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

对于

x^{2} = f (a)

解为

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

u = \frac{1}{4}, u = - \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{1}{2}

使用法则

1 = 1

= \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

- \frac{1}{4}

化简

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

使用法则

1 = 1

= - \frac{1}{2}

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

u = cos (θ)

代回

cos (θ) = \frac{1}{2}, cos (θ) = - \frac{1}{2}

cos (θ) = \frac{1}{2}, cos (θ) = - \frac{1}{2}

cos (θ) = \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{π}{3}, θ = \frac{5 π}{3}

cos (θ) = \frac{1}{2}, 0 \leq θ < 2 π

cos (θ) = \frac{1}{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}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

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

在

0 \leq θ < 2 π

范围内的解

θ = \frac{π}{3}, θ = \frac{5 π}{3}

cos (θ) = - \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

cos (θ) = - \frac{1}{2}, 0 \leq θ < 2 π

cos (θ) = - \frac{1}{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{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

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

在

0 \leq θ < 2 π

范围内的解

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

合并所有解

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

作图

流行的例子

cos(x/2+pi/3)=(sqrt(2))/2

cos (\frac{x}{2} + \frac{π}{3}) = \frac{2}{2}

tan (θ) = 8.2

2sin(x)=4sin(x)+1

2 sin (x) = 4 sin (x) + 1

-1-cos(θ)=-3cos(θ)

- 1 - cos (θ) = - 3 cos (θ)

3*cos(θ)=2-sin(θ)

3 \cdot cos (θ) = 2 - sin (θ)