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受欢迎的三角函数 >

10sin(x)cos^2(x)+3cos(x)=0

解答

10 sin (x) cos^{2} (x) + 3 cos (x) = 0

解答

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = - \frac{0.64350 \dots}{2} + πn, x = \frac{π}{2} + \frac{0.64350 \dots}{2} + πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = - 18.43494 \dots^{\circ} + 18 0^{\circ} n, x = 108.43494 \dots^{\circ} + 18 0^{\circ} n

求解步骤

10 sin (x) cos^{2} (x) + 3 cos (x) = 0

分解

10 sin (x) cos^{2} (x) + 3 cos (x) : cos (x) (10 sin (x) cos (x) + 3)

10 sin (x) cos^{2} (x) + 3 cos (x)

使用指数法则:

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

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

= 10 cos (x) cos (x) + 3 cos (x)

因式分解出通项

cos (x)

= cos (x) (10 sin (x) cos (x) + 3)

cos (x) (10 sin (x) cos (x) + 3) = 0

分别求解每个部分

cos (x) = 0 or 10 sin (x) cos (x) + 3 = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0

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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

10 sin (x) cos (x) + 3 = 0 : x = - \frac{arcsin ( \frac{3}{5} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + πn

10 sin (x) cos (x) + 3 = 0

使用三角恒等式改写

10 sin (x) cos (x) + 3

使用倍角公式:

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

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

= 3 + 10 \cdot \frac{sin ( 2 x )}{2}

3 + 10 \cdot \frac{sin ( 2 x )}{2} = 0

10 \cdot \frac{sin ( 2 x )}{2} = 5 sin (2 x)

10 \cdot \frac{sin ( 2 x )}{2}

分式相乘:

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

= \frac{sin ( 2 x ) \cdot 10}{2}

数字相除:

\frac{10}{2} = 5

= 5 sin (2 x)

3 + 5 sin (2 x) = 0

将

3

到右边

3 + 5 sin (2 x) = 0

两边减去

3

3 + 5 sin (2 x) - 3 = 0 - 3

化简

5 sin (2 x) = - 3

5 sin (2 x) = - 3

两边除以

5

5 sin (2 x) = - 3

两边除以

5

\frac{5 sin ( 2 x )}{5} = \frac{- 3}{5}

化简

sin (2 x) = - \frac{3}{5}

sin (2 x) = - \frac{3}{5}

使用反三角函数性质

sin (2 x) = - \frac{3}{5}

sin (2 x) = - \frac{3}{5}

的通解

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

2 x = arcsin (- \frac{3}{5}) + 2 πn, 2 x = π + arcsin (\frac{3}{5}) + 2 πn

2 x = arcsin (- \frac{3}{5}) + 2 πn, 2 x = π + arcsin (\frac{3}{5}) + 2 πn

解

2 x = arcsin (- \frac{3}{5}) + 2 πn : x = - \frac{arcsin ( \frac{3}{5} )}{2} + πn

2 x = arcsin (- \frac{3}{5}) + 2 πn

化简

arcsin (- \frac{3}{5}) + 2 πn : - arcsin (\frac{3}{5}) + 2 πn

arcsin (- \frac{3}{5}) + 2 πn

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{3}{5}) = - arcsin (\frac{3}{5})

= - arcsin (\frac{3}{5}) + 2 πn

2 x = - arcsin (\frac{3}{5}) + 2 πn

两边除以

2

2 x = - arcsin (\frac{3}{5}) + 2 πn

两边除以

2

\frac{2 x}{2} = - \frac{arcsin ( \frac{3}{5} )}{2} + \frac{2 πn}{2}

化简

x = - \frac{arcsin ( \frac{3}{5} )}{2} + πn

x = - \frac{arcsin ( \frac{3}{5} )}{2} + πn

解

2 x = π + arcsin (\frac{3}{5}) + 2 πn : x = \frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + πn

2 x = π + arcsin (\frac{3}{5}) + 2 πn

两边除以

2

2 x = π + arcsin (\frac{3}{5}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + πn

x = \frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + πn

x = - \frac{arcsin ( \frac{3}{5} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + πn

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = - \frac{arcsin ( \frac{3}{5} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + πn

以小数形式表示解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = - \frac{0.64350 \dots}{2} + πn, x = \frac{π}{2} + \frac{0.64350 \dots}{2} + πn

作图

流行的例子

(2 cos (x)) - 1 = 0

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

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

sin(θ)=(5pi)/3

sin (θ) = \frac{5 π}{3}

10 cos (θ) = 8

cos(θ)=(-18)/(sqrt(20)*\sqrt{26)}

cos (θ) = \frac{- 18}{20 \cdot 26}