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受欢迎的三角函数 >

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

解答

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

解答

x = \frac{π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

3 sin (x)

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

= 1 - sin^{2} (x) - 3 sin (x) + 3 sin^{2} (x)

化简

1 - sin^{2} (x) - 3 sin (x) + 3 sin^{2} (x) : 2 sin^{2} (x) - 3 sin (x) + 1

1 - sin^{2} (x) - 3 sin (x) + 3 sin^{2} (x)

对同类项分组

= - sin^{2} (x) - 3 sin (x) + 3 sin^{2} (x) + 1

同类项相加:

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

= 2 sin^{2} (x) - 3 sin (x) + 1

= 2 sin^{2} (x) - 3 sin (x) + 1

1 + 2 sin^{2} (x) - 3 sin (x) = 0

用替代法求解

1 + 2 sin^{2} (x) - 3 sin (x) = 0

令

: sin (x) = u

1 + 2 u^{2} - 3 u = 0

1 + 2 u^{2} - 3 u = 0 : u = 1, u = \frac{1}{2}

1 + 2 u^{2} - 3 u = 0

改写成标准形式

a x^{2} + b x + c = 0

2 u^{2} - 3 u + 1 = 0

使用求根公式求解

2 u^{2} - 3 u + 1 = 0

二次方程求根公式:

若

a = 2, b = - 3, c = 1

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 \cdot 1 = 1

(- 3)^{2} - 4 \cdot 2 \cdot 1

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数字相减:

9 - 8 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 1}{2 \cdot 2}, u_{2} = \frac{- ( - 3 ) - 1}{2 \cdot 2}

u = \frac{- ( - 3 ) + 1}{2 \cdot 2} : 1

\frac{- ( - 3 ) + 1}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{3 + 1}{2 \cdot 2}

数字相加:

3 + 1 = 4

= \frac{4}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 3 ) - 1}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 3 ) - 1}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{3 - 1}{2 \cdot 2}

数字相减:

3 - 1 = 2

= \frac{2}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

u = 1, u = \frac{1}{2}

u = sin (x)

代回

sin (x) = 1, sin (x) = \frac{1}{2}

sin (x) = 1, sin (x) = \frac{1}{2}

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

sin (x) = 1

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 = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

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 = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

作图

流行的例子

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(\frac{( 1 - sin ^{2} ( x ) )}{( 1 + 2 \cdot sin ^{2} ( x ) )}) = 0.45

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tan(θ+10)=cot(2θ-10)

tan (θ + 1 0^{\circ}) = cot (2 θ - 10)