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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 = 0, x = 2 π

+1

度数

x = 0^{\circ}, x = 36 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

使用指数法则:

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

若

n

是偶数

(- 6)^{2} = 6^{2}

= 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}

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 3 = 6

= \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 = 0, x = 2 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 = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = 0, x = 2 π

合并所有解

x = 0, x = 2 π

作图

流行的例子

csc(θ)+2=0,0<= θ<= 2pi

csc (θ) + 2 = 0, 0 \leq θ \leq 2 π

3sin^2(θ)-sin(θ)-4=0

3 sin^{2} (θ) - sin (θ) - 4 = 0

tan^2(x)=sin(x)sec(x)

tan^{2} (x) = sin (x) sec (x)

sqrt(3)tan(x)-2=-1

3 tan (x) - 2 = - 1

solvefor x,1-cos(x)=0

so l v e f or x, 1 - cos (x) = 0