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

解答

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

解答

x = 2 πn, x = π + 2 πn, x = - 0.95531 \dots + πn, x = 0.95531 \dots + πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = - 54.73561 \dots^{\circ} + 18 0^{\circ} n, x = 54.73561 \dots^{\circ} + 18 0^{\circ} n

求解步骤

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

两边减去

sin (2 x)

sin^{2} (x) tan (x) - sin (2 x) = 0

使用三角恒等式改写

- sin (2 x) + sin^{2} (x) tan (x)

使用倍角公式:

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

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

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

分解

sin^{2} (x) tan (x) - 2 cos (x) sin (x) : sin (x) (sin (x) tan (x) - 2 cos (x))

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

因式分解出通项

sin (x)

= sin (x) (sin (x) tan (x) - 2 cos (x))

sin (x) (sin (x) tan (x) - 2 cos (x)) = 0

分别求解每个部分

sin (x) = 0 or sin (x) tan (x) - 2 cos (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 0

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 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, x = π + 2 πn

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

sin (x) tan (x) - 2 cos (x) = 0

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

化简

- 2 cos (x) + sin (x) \frac{sin ( x )}{cos ( x )} : \frac{- 2 cos ^{2} ( x ) + sin ^{2} ( x )}{cos ( x )}

- 2 cos (x) + sin (x) \frac{sin ( x )}{cos ( x )}

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

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

分式相乘:

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

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

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

sin (x) sin (x)

使用指数法则:

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

sin (x) sin (x) = sin^{1 + 1} (x)

= sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= sin^{2} (x)

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

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

将项转换为分式:

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

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

因为分母相等，所以合并分式:

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

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

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

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

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

2 cos (x) cos (x)

使用指数法则:

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

cos (x) cos (x) = cos^{1 + 1} (x)

= 2 cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= 2 cos^{2} (x)

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

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

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

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

分解

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

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

将

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

改写为

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

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

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

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

使用指数法则:

a^{m} b^{m} = (ab)^{m}

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

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

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

分别求解每个部分

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

两边加上

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

合并所有解

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

合并所有解

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

以小数形式表示解

x = 2 πn, x = π + 2 πn, x = - 0.95531 \dots + πn, x = 0.95531 \dots + πn

流行的例子

6-9sin(θ)-4cos^2(θ)=0

6 - 9 sin (θ) - 4 cos^{2} (θ) = 0

2cos^2(x)-1cos(x)=0

2 cos^{2} (x) - 1 cos (x) = 0

sec (x) = - 1.5

sin(x)+2=-csc(x)

sin (x) + 2 = - csc (x)

cos (a) = \frac{9}{14}