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受欢迎的三角函数 >

8-12sin^2(x)=4cos^2(x)

解答

8 - 12 sin^{2} (x) = 4 cos^{2} (x)

解答

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

+1

度数

x = 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

求解步骤

8 - 12 sin^{2} (x) = 4 cos^{2} (x)

两边减去

4 cos^{2} (x)

8 - 12 sin^{2} (x) - 4 cos^{2} (x) = 0

使用三角恒等式改写

8 - 12 sin^{2} (x) - 4 cos^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= 8 - 12 sin^{2} (x) - 4 (1 - sin^{2} (x))

化简

8 - 12 sin^{2} (x) - 4 (1 - sin^{2} (x)) : - 8 sin^{2} (x) + 4

8 - 12 sin^{2} (x) - 4 (1 - sin^{2} (x))

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (x)

= - 4 \cdot 1 - (- 4) sin^{2} (x)

使用加减运算法则

- (- a) = a

= - 4 \cdot 1 + 4 sin^{2} (x)

数字相乘:

4 \cdot 1 = 4

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

= 8 - 12 sin^{2} (x) - 4 + 4 sin^{2} (x)

化简

8 - 12 sin^{2} (x) - 4 + 4 sin^{2} (x) : - 8 sin^{2} (x) + 4

8 - 12 sin^{2} (x) - 4 + 4 sin^{2} (x)

对同类项分组

= - 12 sin^{2} (x) + 4 sin^{2} (x) + 8 - 4

同类项相加:

- 12 sin^{2} (x) + 4 sin^{2} (x) = - 8 sin^{2} (x)

= - 8 sin^{2} (x) + 8 - 4

数字相加/相减:

8 - 4 = 4

= - 8 sin^{2} (x) + 4

= - 8 sin^{2} (x) + 4

= - 8 sin^{2} (x) + 4

4 - 8 sin^{2} (x) = 0

用替代法求解

4 - 8 sin^{2} (x) = 0

令

: sin (x) = u

4 - 8 u^{2} = 0

4 - 8 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

4 - 8 u^{2} = 0

将

4

到右边

4 - 8 u^{2} = 0

两边减去

4

4 - 8 u^{2} - 4 = 0 - 4

化简

- 8 u^{2} = - 4

- 8 u^{2} = - 4

两边除以

- 8

- 8 u^{2} = - 4

两边除以

- 8

\frac{- 8 u ^{2}}{- 8} = \frac{- 4}{- 8}

化简

\frac{- 8 u ^{2}}{- 8} = \frac{- 4}{- 8}

化简

\frac{- 8 u ^{2}}{- 8} : u^{2}

\frac{- 8 u ^{2}}{- 8}

使用分式法则:

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

= \frac{8 u ^{2}}{8}

数字相除:

\frac{8}{8} = 1

= u^{2}

化简

\frac{- 4}{- 8} : \frac{1}{2}

\frac{- 4}{- 8}

使用分式法则:

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

= \frac{4}{8}

约分:

4

= \frac{1}{2}

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

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

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

对于

x^{2} = f (a)

解为

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

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

u = sin (x)

代回

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

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

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

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

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

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

sin (x) = - \frac{1}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

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

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

合并所有解

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

作图

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