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受欢迎的三角函数 >

sin^2(θ)+3cos(2θ)=2

解答

sin^{2} (θ) + 3 cos (2 θ) = 2

解答

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

+1

度数

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

求解步骤

sin^{2} (θ) + 3 cos (2 θ) = 2

两边减去

2

sin^{2} (θ) + 3 cos (2 θ) - 2 = 0

使用三角恒等式改写

- 2 + sin^{2} (θ) + 3 cos (2 θ)

使用倍角公式:

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

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

化简

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

- 2 + sin^{2} (θ) + 3 (1 - 2 sin^{2} (θ))

乘开

3 (1 - 2 sin^{2} (θ)) : 3 - 6 sin^{2} (θ)

3 (1 - 2 sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = 2 sin^{2} (θ)

= 3 \cdot 1 - 3 \cdot 2 sin^{2} (θ)

化简

3 \cdot 1 - 3 \cdot 2 sin^{2} (θ) : 3 - 6 sin^{2} (θ)

3 \cdot 1 - 3 \cdot 2 sin^{2} (θ)

数字相乘:

3 \cdot 1 = 3

= 3 - 3 \cdot 2 sin^{2} (θ)

数字相乘:

3 \cdot 2 = 6

= 3 - 6 sin^{2} (θ)

= 3 - 6 sin^{2} (θ)

= - 2 + sin^{2} (θ) + 3 - 6 sin^{2} (θ)

化简

- 2 + sin^{2} (θ) + 3 - 6 sin^{2} (θ) : - 5 sin^{2} (θ) + 1

- 2 + sin^{2} (θ) + 3 - 6 sin^{2} (θ)

对同类项分组

= sin^{2} (θ) - 6 sin^{2} (θ) - 2 + 3

同类项相加:

sin^{2} (θ) - 6 sin^{2} (θ) = - 5 sin^{2} (θ)

= - 5 sin^{2} (θ) - 2 + 3

数字相加/相减:

- 2 + 3 = 1

= - 5 sin^{2} (θ) + 1

= - 5 sin^{2} (θ) + 1

= - 5 sin^{2} (θ) + 1

1 - 5 sin^{2} (θ) = 0

用替代法求解

1 - 5 sin^{2} (θ) = 0

令

: sin (θ) = 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 (θ)

代回

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

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

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

sin (θ) = \frac{1}{5}

使用反三角函数性质

sin (θ) = \frac{1}{5}

sin (θ) = \frac{1}{5}

的通解

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

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

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

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

sin (θ) = - \frac{1}{5}

使用反三角函数性质

sin (θ) = - \frac{1}{5}

sin (θ) = - \frac{1}{5}

的通解

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

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

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

合并所有解

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

以小数形式表示解

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

作图

流行的例子

cos (x) = 0.48

cos (x) = 0.34

cos (x) = 0.39

1-sqrt(2)sin(x)=cos(2x)

1 - 2 sin (x) = cos (2 x)

cos (x) = 0.89