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受欢迎的三角函数 >

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

解答

4 sin^{2} (x) + 4 cos (2 x) = 1

解答

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

+1

度数

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

求解步骤

4 sin^{2} (x) + 4 cos (2 x) = 1

两边减去

1

4 sin^{2} (x) + 4 cos (2 x) - 1 = 0

使用三角恒等式改写

- 1 + 4 cos (2 x) + 4 sin^{2} (x)

使用倍角公式:

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

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

= - 1 + 4 cos (2 x) - 4 (cos (2 x) - cos^{2} (x))

化简

- 1 + 4 cos (2 x) - 4 (cos (2 x) - cos^{2} (x)) : - 1 + 4 cos^{2} (x)

- 1 + 4 cos (2 x) - 4 (cos (2 x) - cos^{2} (x))

乘开

- 4 (cos (2 x) - cos^{2} (x)) : - 4 cos (2 x) + 4 cos^{2} (x)

- 4 (cos (2 x) - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 4, b = cos (2 x), c = cos^{2} (x)

= - 4 cos (2 x) - (- 4) cos^{2} (x)

使用加减运算法则

- (- a) = a

= - 4 cos (2 x) + 4 cos^{2} (x)

= - 1 + 4 cos (2 x) - 4 cos (2 x) + 4 cos^{2} (x)

同类项相加:

4 cos (2 x) - 4 cos (2 x) = 0

= - 1 + 4 cos^{2} (x)

= - 1 + 4 cos^{2} (x)

- 1 + 4 cos^{2} (x) = 0

用替代法求解

- 1 + 4 cos^{2} (x) = 0

令

: cos (x) = u

- 1 + 4 u^{2} = 0

- 1 + 4 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

- 1 + 4 u^{2} = 0

将

1

到右边

- 1 + 4 u^{2} = 0

两边加上

1

- 1 + 4 u^{2} + 1 = 0 + 1

化简

4 u^{2} = 1

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:

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:

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:

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:

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 (x)

代回

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

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

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

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

cos (x) = \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}

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

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

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

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

cos (x) = - \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}

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

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

合并所有解

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

作图

流行的例子

cot(θ)=4,cos(θ)<0

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