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受欢迎的三角函数 >

4(cos(x))^2=5-4sin(x)

解答

4 (cos (x))^{2} = 5 - 4 sin (x)

解答

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

+1

度数

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

求解步骤

4 (cos (x))^{2} = 5 - 4 sin (x)

两边减去

5 - 4 sin (x)

4 cos^{2} (x) - 5 + 4 sin (x) = 0

使用三角恒等式改写

- 5 + 4 cos^{2} (x) + 4 sin (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)

= - 5 + 4 (1 - sin^{2} (x)) + 4 sin (x)

化简

- 5 + 4 (1 - sin^{2} (x)) + 4 sin (x) : 4 sin (x) - 4 sin^{2} (x) - 1

- 5 + 4 (1 - sin^{2} (x)) + 4 sin (x)

乘开

4 (1 - sin^{2} (x)) : 4 - 4 sin^{2} (x)

4 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

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

数字相乘:

4 \cdot 1 = 4

= 4 - 4 sin^{2} (x)

= - 5 + 4 - 4 sin^{2} (x) + 4 sin (x)

数字相加/相减:

- 5 + 4 = - 1

= 4 sin (x) - 4 sin^{2} (x) - 1

= 4 sin (x) - 4 sin^{2} (x) - 1

- 1 + 4 sin (x) - 4 sin^{2} (x) = 0

用替代法求解

- 1 + 4 sin (x) - 4 sin^{2} (x) = 0

令

: sin (x) = u

- 1 + 4 u - 4 u^{2} = 0

- 1 + 4 u - 4 u^{2} = 0 : u = \frac{1}{2}

- 1 + 4 u - 4 u^{2} = 0

改写成标准形式

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

- 4 u^{2} + 4 u - 1 = 0

使用求根公式求解

- 4 u^{2} + 4 u - 1 = 0

二次方程求根公式:

若

a = - 4, b = 4, c = - 1

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

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

4^{2} - 4 (- 4) (- 1) = 0

4^{2} - 4 (- 4) (- 1)

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

数字相减:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- 4 \pm 0}{2 ( - 4 )}

u = \frac{- 4}{2 ( - 4 )}

\frac{- 4}{2 ( - 4 )} = \frac{1}{2}

\frac{- 4}{2 ( - 4 )}

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

使用分式法则:

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

= \frac{4}{8}

约分:

4

= \frac{1}{2}

u = \frac{1}{2}

二次方程组的解是:

u = \frac{1}{2}

u = sin (x)

代回

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

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

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

作图

流行的例子

tan (2 t) = 1

21=3*2*25sin(x)+9.8*3*2*3*2*0.5

21 = 3 \cdot 2 \cdot 25 sin (x) + 9.8 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 0.5

sin(2x)=((3m-8))/6

sin (2 x) = \frac{( 3 m - 8 )}{6}

cos(x)(18-12sqrt(2))=12+6sqrt(3)

cos (x) (18 - 122) = 12 + 63

cos(2x+pi/3)=(sqrt(3))/2 ,0<= x<= 2pi

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