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受欢迎的三角函数 >

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

解答

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

解答

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

+1

度数

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

求解步骤

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

两边减去

2 cos^{2} (t)

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

- (1 - sin^{2} (t)) \cdot 2 + 3 sin (t) = 0

用替代法求解

- (1 - sin^{2} (t)) \cdot 2 + 3 sin (t) = 0

令

: sin (t) = u

- (1 - u^{2}) \cdot 2 + 3 u = 0

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

- (1 - u^{2}) \cdot 2 + 3 u = 0

展开

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

- (1 - u^{2}) \cdot 2 + 3 u

= - 2 (1 - u^{2}) + 3 u

乘开

- 2 (1 - u^{2}) : - 2 + 2 u^{2}

- 2 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = - 2, b = 1, c = u^{2}

= - 2 \cdot 1 - (- 2) u^{2}

使用加减运算法则

- (- a) = a

= - 2 \cdot 1 + 2 u^{2}

数字相乘:

2 \cdot 1 = 2

= - 2 + 2 u^{2}

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

- 2 + 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 - 2 = 0

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

3^{2} - 4 \cdot 2 (- 2) = 5

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

使用法则

- (- a) = a

= 3^{2} + 4 \cdot 2 \cdot 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

数字相加:

9 + 16 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- 3 + 5}{2 \cdot 2}, u_{2} = \frac{- 3 - 5}{2 \cdot 2}

u = \frac{- 3 + 5}{2 \cdot 2} : \frac{1}{2}

\frac{- 3 + 5}{2 \cdot 2}

数字相加/相减:

- 3 + 5 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

u = \frac{- 3 - 5}{2 \cdot 2} : - 2

\frac{- 3 - 5}{2 \cdot 2}

数字相减:

- 3 - 5 = - 8

= \frac{- 8}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 8}{4}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= - 2

二次方程组的解是:

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

u = sin (t)

代回

sin (t) = \frac{1}{2}, sin (t) = - 2

sin (t) = \frac{1}{2}, sin (t) = - 2

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

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

sin (t) = \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}

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

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

sin (t) = - 2 :

无解

sin (t) = - 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

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

作图

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