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受欢迎的三角函数 >

10sin^2(x)=10+5cos(x)

解答

10 sin^{2} (x) = 10 + 5 cos (x)

解答

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

+1

度数

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

10 sin^{2} (x) = 10 + 5 cos (x)

两边减去

10 + 5 cos (x)

10 sin^{2} (x) - 10 - 5 cos (x) = 0

使用三角恒等式改写

- 10 + 10 sin^{2} (x) - 5 cos (x)

使用毕达哥拉斯恒等式:

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

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

= - 10 + 10 (1 - cos^{2} (x)) - 5 cos (x)

化简

- 10 + 10 (1 - cos^{2} (x)) - 5 cos (x) : - 10 cos^{2} (x) - 5 cos (x)

- 10 + 10 (1 - cos^{2} (x)) - 5 cos (x)

乘开

10 (1 - cos^{2} (x)) : 10 - 10 cos^{2} (x)

10 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 10, b = 1, c = cos^{2} (x)

= 10 \cdot 1 - 10 cos^{2} (x)

数字相乘:

10 \cdot 1 = 10

= 10 - 10 cos^{2} (x)

= - 10 + 10 - 10 cos^{2} (x) - 5 cos (x)

- 10 + 10 = 0

= - 10 cos^{2} (x) - 5 cos (x)

= - 10 cos^{2} (x) - 5 cos (x)

- 10 cos^{2} (x) - 5 cos (x) = 0

用替代法求解

- 10 cos^{2} (x) - 5 cos (x) = 0

令

: cos (x) = u

- 10 u^{2} - 5 u = 0

- 10 u^{2} - 5 u = 0 : u = - \frac{1}{2}, u = 0

- 10 u^{2} - 5 u = 0

使用求根公式求解

- 10 u^{2} - 5 u = 0

二次方程求根公式:

若

a = - 10, b = - 5, c = 0

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 10 ) \cdot 0}{2 ( - 10 )}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 10 ) \cdot 0}{2 ( - 10 )}

(- 5)^{2} - 4 (- 10) \cdot 0 = 5

(- 5)^{2} - 4 (- 10) \cdot 0

使用法则

- (- a) = a

= (- 5)^{2} + 4 \cdot 10 \cdot 0

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 10 \cdot 0

使用法则

0 \cdot a = 0

= 5^{2} + 0

5^{2} + 0 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 5

u_{1, 2} = \frac{- ( - 5 ) \pm 5}{2 ( - 10 )}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 5}{2 ( - 10 )}, u_{2} = \frac{- ( - 5 ) - 5}{2 ( - 10 )}

u = \frac{- ( - 5 ) + 5}{2 ( - 10 )} : - \frac{1}{2}

\frac{- ( - 5 ) + 5}{2 ( - 10 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{5 + 5}{- 2 \cdot 10}

数字相加:

5 + 5 = 10

= \frac{10}{- 2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{10}{- 20}

使用分式法则:

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

= - \frac{10}{20}

约分:

10

= - \frac{1}{2}

u = \frac{- ( - 5 ) - 5}{2 ( - 10 )} : 0

\frac{- ( - 5 ) - 5}{2 ( - 10 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{5 - 5}{- 2 \cdot 10}

数字相减:

5 - 5 = 0

= \frac{0}{- 2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{0}{- 20}

使用分式法则:

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

= - \frac{0}{20}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

二次方程组的解是:

u = - \frac{1}{2}, u = 0

u = cos (x)

代回

cos (x) = - \frac{1}{2}, cos (x) = 0

cos (x) = - \frac{1}{2}, cos (x) = 0

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = - \frac{1}{2}

cos (x) = - \frac{1}{2}

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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

合并所有解

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

作图

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