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受欢迎的三角函数 >

sin^2(x)-sin(x)cos(x)-6cos^2(x)=0

解答

sin^{2} (x) - sin (x) cos (x) - 6 cos^{2} (x) = 0

解答

x = - 1.10714 \dots + πn, x = 1.24904 \dots + πn

+1

度数

x = - 63.43494 \dots^{\circ} + 18 0^{\circ} n, x = 71.56505 \dots^{\circ} + 18 0^{\circ} n

求解步骤

sin^{2} (x) - sin (x) cos (x) - 6 cos^{2} (x) = 0

分解

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

sin^{2} (x) - sin (x) cos (x) - 6 cos^{2} (x)

将表达式拆分成组

sin^{2} (x) - sin (x) cos (x) - 6 cos^{2} (x)

定义

6

的因数:

1, 2, 3, 6

6

约数 (因数)

找到

6

的质因数:

2, 3

6

6

除以

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3

添加质因数:

2, 3

将 1 和数字

6

自身相加

1, 6

6

的因数

1, 2, 3, 6

6

的负因数:

- 1, - 2, - 3, - 6

将因数乘以

- 1

得到负因数

- 1, - 2, - 3, - 6

对于每两个因数

u * v = - 6

，检验是否

u + v = - 1

检验

u = 1, v = - 6 : u * v = - 6, u + v = - 5 \Rightarrow

假检验

u = 2, v = - 3 : u * v = - 6, u + v = - 1 \Rightarrow

真

u = 2, v = - 3

分组为

(a x^{2} + ux y) + (vx y + c y^{2})

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

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

从

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

分解出因式

sin (x)

:

sin (x) (sin (x) + 2 cos (x))

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

使用指数法则:

a^{b + c} = a^{b} a^{c}

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

= sin (x) sin (x) + 2 sin (x) cos (x)

因式分解出通项

sin (x)

= sin (x) (sin (x) + 2 cos (x))

从

- 3 sin (x) cos (x) - 6 cos^{2} (x)

分解出因式

- 3 cos (x)

:

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

- 3 sin (x) cos (x) - 6 cos^{2} (x)

使用指数法则:

a^{b + c} = a^{b} a^{c}

cos^{2} (x) = cos (x) cos (x)

= - 3 sin (x) cos (x) - 6 cos (x) cos (x)

将

- 6

改写为

2 \cdot 3

= - 3 sin (x) cos (x) + 2 \cdot 3 cos (x) cos (x)

因式分解出通项

- 3 cos (x)

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

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

因式分解出通项

sin (x) + 2 cos (x)

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

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

分别求解每个部分

sin (x) + 2 cos (x) = 0 or sin (x) - 3 cos (x) = 0

sin (x) + 2 cos (x) = 0 : x = arctan (- 2) + πn

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

使用三角恒等式改写

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

在两边除以

cos (x), cos (x) \neq = 0

\frac{sin ( x ) + 2 cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

化简

\frac{sin ( x )}{cos ( x )} + 2 = 0

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

tan (x) + 2 = 0

tan (x) + 2 = 0

将

2

到右边

tan (x) + 2 = 0

两边减去

2

tan (x) + 2 - 2 = 0 - 2

化简

tan (x) = - 2

tan (x) = - 2

使用反三角函数性质

tan (x) = - 2

tan (x) = - 2

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- 2) + πn

x = arctan (- 2) + πn

sin (x) - 3 cos (x) = 0 : x = arctan (3) + πn

sin (x) - 3 cos (x) = 0

使用三角恒等式改写

sin (x) - 3 cos (x) = 0

在两边除以

cos (x), cos (x) \neq = 0

\frac{sin ( x ) - 3 cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

化简

\frac{sin ( x )}{cos ( x )} - 3 = 0

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

tan (x) - 3 = 0

tan (x) - 3 = 0

将

3

到右边

tan (x) - 3 = 0

两边加上

3

tan (x) - 3 + 3 = 0 + 3

化简

tan (x) = 3

tan (x) = 3

使用反三角函数性质

tan (x) = 3

tan (x) = 3

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (3) + πn

x = arctan (3) + πn

合并所有解

x = arctan (- 2) + πn, x = arctan (3) + πn

以小数形式表示解

x = - 1.10714 \dots + πn, x = 1.24904 \dots + πn

作图

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