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受欢迎的三角函数 >

2cos^2(θ)-3sin(θ)+2=0

解答

2 cos^{2} (θ) - 3 sin (θ) + 2 = 0

解答

θ = 1.01746 \dots + 2 πn, θ = π - 1.01746 \dots + 2 πn

+1

度数

θ = 58.29672 \dots^{\circ} + 36 0^{\circ} n, θ = 121.70327 \dots^{\circ} + 36 0^{\circ} n

求解步骤

2 cos^{2} (θ) - 3 sin (θ) + 2 = 0

使用三角恒等式改写

2 + 2 cos^{2} (θ) - 3 sin (θ)

使用毕达哥拉斯恒等式:

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

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

= 2 + 2 (1 - sin^{2} (θ)) - 3 sin (θ)

化简

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

2 + 2 (1 - sin^{2} (θ)) - 3 sin (θ)

乘开

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

2 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (θ)

= 2 \cdot 1 - 2 sin^{2} (θ)

数字相乘:

2 \cdot 1 = 2

= 2 - 2 sin^{2} (θ)

= 2 + 2 - 2 sin^{2} (θ) - 3 sin (θ)

数字相加:

2 + 2 = 4

= - 2 sin^{2} (θ) - 3 sin (θ) + 4

= - 2 sin^{2} (θ) - 3 sin (θ) + 4

4 - 2 sin^{2} (θ) - 3 sin (θ) = 0

用替代法求解

4 - 2 sin^{2} (θ) - 3 sin (θ) = 0

令

: sin (θ) = u

4 - 2 u^{2} - 3 u = 0

4 - 2 u^{2} - 3 u = 0 : u = - \frac{3 + 41}{4}, u = \frac{41 - 3}{4}

4 - 2 u^{2} - 3 u = 0

改写成标准形式

a x^{2} + b x + c = 0

- 2 u^{2} - 3 u + 4 = 0

使用求根公式求解

- 2 u^{2} - 3 u + 4 = 0

二次方程求根公式:

若

a = - 2, b = - 3, c = 4

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) \cdot 4}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) \cdot 4}{2 ( - 2 )}

(- 3)^{2} - 4 (- 2) \cdot 4 = 41

(- 3)^{2} - 4 (- 2) \cdot 4

使用法则

- (- a) = a

= (- 3)^{2} + 4 \cdot 2 \cdot 4

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 2 \cdot 4

数字相乘:

4 \cdot 2 \cdot 4 = 32

= 3^{2} + 32

3^{2} = 9

= 9 + 32

数字相加:

9 + 32 = 41

= 41

u_{1, 2} = \frac{- ( - 3 ) \pm 41}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 41}{2 ( - 2 )}, u_{2} = \frac{- ( - 3 ) - 41}{2 ( - 2 )}

u = \frac{- ( - 3 ) + 41}{2 ( - 2 )} : - \frac{3 + 41}{4}

\frac{- ( - 3 ) + 41}{2 ( - 2 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{3 + 41}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 + 41}{- 4}

使用分式法则:

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

= - \frac{3 + 41}{4}

u = \frac{- ( - 3 ) - 41}{2 ( - 2 )} : \frac{41 - 3}{4}

\frac{- ( - 3 ) - 41}{2 ( - 2 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{3 - 41}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 - 41}{- 4}

使用分式法则:

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

3 - 41 = - (41 - 3)

= \frac{41 - 3}{4}

二次方程组的解是:

u = - \frac{3 + 41}{4}, u = \frac{41 - 3}{4}

u = sin (θ)

代回

sin (θ) = - \frac{3 + 41}{4}, sin (θ) = \frac{41 - 3}{4}

sin (θ) = - \frac{3 + 41}{4}, sin (θ) = \frac{41 - 3}{4}

sin (θ) = - \frac{3 + 41}{4} :

无解

sin (θ) = - \frac{3 + 41}{4}

- 1 \leq sin (x) \leq 1

无解

sin (θ) = \frac{41 - 3}{4} : θ = arcsin (\frac{41 - 3}{4}) + 2 πn, θ = π - arcsin (\frac{41 - 3}{4}) + 2 πn

sin (θ) = \frac{41 - 3}{4}

使用反三角函数性质

sin (θ) = \frac{41 - 3}{4}

sin (θ) = \frac{41 - 3}{4}

的通解

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

θ = arcsin (\frac{41 - 3}{4}) + 2 πn, θ = π - arcsin (\frac{41 - 3}{4}) + 2 πn

θ = arcsin (\frac{41 - 3}{4}) + 2 πn, θ = π - arcsin (\frac{41 - 3}{4}) + 2 πn

合并所有解

θ = arcsin (\frac{41 - 3}{4}) + 2 πn, θ = π - arcsin (\frac{41 - 3}{4}) + 2 πn

以小数形式表示解

θ = 1.01746 \dots + 2 πn, θ = π - 1.01746 \dots + 2 πn

作图

流行的例子

sin(x)+sin(2x)+sin(3x)=0

sin (x) + sin (2 x) + sin (3 x) = 0

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

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

36 cos^{2} (x) - 9 = 0

sin (x) = \frac{5}{6}

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

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