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受欢迎的三角函数 >

8cos^2(3x)+6cos(6x)=10216

解答

8 \cdot cos^{2} (3 x) + 6 \cdot cos (6 x) = 10216

解答

x \in R 无解

求解步骤

8 cos^{2} (3 x) + 6 cos (6 x) = 10216

两边减去

10216

8 cos^{2} (3 x) + 6 cos (6 x) - 10216 = 0

使用三角恒等式改写

- 10216 + 6 cos (6 x) + 8 cos^{2} (3 x)

利用以下特性:

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

使用倍角公式

cos (2 θ) = 2 cos^{2} (θ) - 1

交换两边

2 cos^{2} (θ) - 1 = cos (2 θ)

两边加上

1

2 sin^{2} (θ) = 1 + cos (2 θ)

两边除以

2

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

= - 10216 + 6 cos (6 x) + 8 \cdot \frac{1 + cos ( 2 \cdot 3 x )}{2}

化简

- 10216 + 6 cos (6 x) + 8 \cdot \frac{1 + cos ( 2 \cdot 3 x )}{2} : 10 cos (6 x) - 10212

- 10216 + 6 cos (6 x) + 8 \cdot \frac{1 + cos ( 2 \cdot 3 x )}{2}

8 \cdot \frac{1 + cos ( 2 \cdot 3 x )}{2} = 4 (cos (6 x) + 1)

8 \cdot \frac{1 + cos ( 2 \cdot 3 x )}{2}

数字相乘:

2 \cdot 3 = 6

= 8 \cdot \frac{cos ( 6 x ) + 1}{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 1 + cos ( 6 x ) ) \cdot 8}{2}

数字相除:

\frac{8}{2} = 4

= 4 (cos (6 x) + 1)

= - 10216 + 6 cos (6 x) + 4 (cos (6 x) + 1)

乘开

4 (cos (6 x) + 1) : 4 cos (6 x) + 4

4 (cos (6 x) + 1)

使用分配律:

a (b + c) = ab + a c

a = 4, b = cos (6 x), c = 1

= 4 cos (6 x) + 4 \cdot 1

数字相乘:

4 \cdot 1 = 4

= 4 cos (6 x) + 4

= - 10216 + 6 cos (6 x) + 4 cos (6 x) + 4

化简

- 10216 + 6 cos (6 x) + 4 cos (6 x) + 4 : 10 cos (6 x) - 10212

- 10216 + 6 cos (6 x) + 4 cos (6 x) + 4

同类项相加:

6 cos (6 x) + 4 cos (6 x) = 10 cos (6 x)

= - 10216 + 10 cos (6 x) + 4

对同类项分组

= 10 cos (6 x) - 10216 + 4

数字相加/相减:

- 10216 + 4 = - 10212

= 10 cos (6 x) - 10212

= 10 cos (6 x) - 10212

= 10 cos (6 x) - 10212

- 10212 + 10 cos (6 x) = 0

将

10212

到右边

- 10212 + 10 cos (6 x) = 0

两边加上

10212

- 10212 + 10 cos (6 x) + 10212 = 0 + 10212

化简

10 cos (6 x) = 10212

10 cos (6 x) = 10212

两边除以

10

10 cos (6 x) = 10212

两边除以

10

\frac{10 cos ( 6 x )}{10} = \frac{10212}{10}

化简

cos (6 x) = \frac{5106}{5}

cos (6 x) = \frac{5106}{5}

- 1 \leq cos (x) \leq 1

x \in R 无解

作图

流行的例子

sin(x)=(48*sin(69))/(47.5)

sin (x) = \frac{48 \cdot sin ( 6 9 ^{\circ} )}{47.5}

cos(x)=(8sqrt(3))/(16)

cos (x) = \frac{8 3}{16}

solvefor x,cos^2(x)=0

so l v e f or x, cos^{2} (x) = 0

cos(θ)=(sqrt(7))/4

cos (θ) = \frac{7}{4}

sin(2θ-pi/2)=-1,0<= θ<= 2pi

sin (2 θ - \frac{π}{2}) = - 1, 0 \leq θ \leq 2 π