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受欢迎的三角函数 >

3cos^2(x)+6cos(x)=-3,0<= x<= 2pi

解答

3 cos^{2} (x) + 6 cos (x) = - 3, 0 \leq x \leq 2 π

解答

x = π

+1

度数

x = 18 0^{\circ}

求解步骤

3 cos^{2} (x) + 6 cos (x) = - 3, 0 \leq x \leq 2 π

用替代法求解

3 cos^{2} (x) + 6 cos (x) = - 3

令

: cos (x) = u

3 u^{2} + 6 u = - 3

3 u^{2} + 6 u = - 3 : u = - 1

3 u^{2} + 6 u = - 3

将

3

para o lado esquerdo

3 u^{2} + 6 u = - 3

两边加上

3

3 u^{2} + 6 u + 3 = - 3 + 3

化简

3 u^{2} + 6 u + 3 = 0

3 u^{2} + 6 u + 3 = 0

使用求根公式求解

3 u^{2} + 6 u + 3 = 0

二次方程求根公式:

若

a = 3, b = 6, c = 3

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

6^{2} - 4 \cdot 3 \cdot 3 = 0

6^{2} - 4 \cdot 3 \cdot 3

数字相乘:

4 \cdot 3 \cdot 3 = 36

= 6^{2} - 36

6^{2} = 36

= 36 - 36

数字相减:

36 - 36 = 0

= 0

u_{1, 2} = \frac{- 6 \pm 0}{2 \cdot 3}

u = \frac{- 6}{2 \cdot 3}

\frac{- 6}{2 \cdot 3} = - 1

\frac{- 6}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 6}{6}

使用分式法则:

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

= - \frac{6}{6}

使用法则

\frac{a}{a} = 1

= - 1

u = - 1

二次方程组的解是:

u = - 1

u = cos (x)

代回

cos (x) = - 1

cos (x) = - 1

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

cos (x) = - 1, 0 \leq x \leq 2 π

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = π

合并所有解

x = π

作图

流行的例子

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tan (θ) = 7, sec^{2} (θ)

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5sqrt(3)tan(x)+3=8sqrt(3)tan(x)

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