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受欢迎的三角函数 >

4cos^2(x)=sin^2(x)+3

解答

4 cos^{2} (x) = sin^{2} (x) + 3

解答

x = 0.46364 \dots + 2 πn, x = π - 0.46364 \dots + 2 πn, x = - 0.46364 \dots + 2 πn, x = π + 0.46364 \dots + 2 πn

+1

度数

x = 26.56505 \dots^{\circ} + 36 0^{\circ} n, x = 153.43494 \dots^{\circ} + 36 0^{\circ} n, x = - 26.56505 \dots^{\circ} + 36 0^{\circ} n, x = 206.56505 \dots^{\circ} + 36 0^{\circ} n

求解步骤

4 cos^{2} (x) = sin^{2} (x) + 3

两边减去

sin^{2} (x) + 3

4 cos^{2} (x) - sin^{2} (x) - 3 = 0

使用三角恒等式改写

- 3 - 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)

= - 3 - sin^{2} (x) + 4 (1 - sin^{2} (x))

化简

- 3 - sin^{2} (x) + 4 (1 - sin^{2} (x)) : - 5 sin^{2} (x) + 1

- 3 - 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)

数字相乘:

4 \cdot 1 = 4

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

= - 3 - sin^{2} (x) + 4 - 4 sin^{2} (x)

化简

- 3 - sin^{2} (x) + 4 - 4 sin^{2} (x) : - 5 sin^{2} (x) + 1

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

对同类项分组

= - sin^{2} (x) - 4 sin^{2} (x) - 3 + 4

同类项相加:

- sin^{2} (x) - 4 sin^{2} (x) = - 5 sin^{2} (x)

= - 5 sin^{2} (x) - 3 + 4

数字相加/相减:

- 3 + 4 = 1

= - 5 sin^{2} (x) + 1

= - 5 sin^{2} (x) + 1

= - 5 sin^{2} (x) + 1

1 - 5 sin^{2} (x) = 0

用替代法求解

1 - 5 sin^{2} (x) = 0

令

: sin (x) = u

1 - 5 u^{2} = 0

1 - 5 u^{2} = 0 : u = \frac{1}{5}, u = - \frac{1}{5}

1 - 5 u^{2} = 0

将

1

到右边

1 - 5 u^{2} = 0

两边减去

1

1 - 5 u^{2} - 1 = 0 - 1

化简

- 5 u^{2} = - 1

- 5 u^{2} = - 1

两边除以

- 5

- 5 u^{2} = - 1

两边除以

- 5

\frac{- 5 u ^{2}}{- 5} = \frac{- 1}{- 5}

化简

u^{2} = \frac{1}{5}

u^{2} = \frac{1}{5}

对于

x^{2} = f (a)

解为

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

u = \frac{1}{5}, u = - \frac{1}{5}

u = sin (x)

代回

sin (x) = \frac{1}{5}, sin (x) = - \frac{1}{5}

sin (x) = \frac{1}{5}, sin (x) = - \frac{1}{5}

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

sin (x) = \frac{1}{5}

使用反三角函数性质

sin (x) = \frac{1}{5}

sin (x) = \frac{1}{5}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{1}{5}) + 2 πn, x = π - arcsin (\frac{1}{5}) + 2 πn

x = arcsin (\frac{1}{5}) + 2 πn, x = π - arcsin (\frac{1}{5}) + 2 πn

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

sin (x) = - \frac{1}{5}

使用反三角函数性质

sin (x) = - \frac{1}{5}

sin (x) = - \frac{1}{5}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{1}{5}) + 2 πn, x = π + arcsin (\frac{1}{5}) + 2 πn

x = arcsin (- \frac{1}{5}) + 2 πn, x = π + arcsin (\frac{1}{5}) + 2 πn

合并所有解

x = arcsin (\frac{1}{5}) + 2 πn, x = π - arcsin (\frac{1}{5}) + 2 πn, x = arcsin (- \frac{1}{5}) + 2 πn, x = π + arcsin (\frac{1}{5}) + 2 πn

以小数形式表示解

x = 0.46364 \dots + 2 πn, x = π - 0.46364 \dots + 2 πn, x = - 0.46364 \dots + 2 πn, x = π + 0.46364 \dots + 2 πn

作图

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