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受欢迎的三角函数 >

4sin^2(x)=5-4cos(x),0<= x<2pi

解答

4 sin^{2} (x) = 5 - 4 cos (x), 0 \leq x < 2 π

解答

x = \frac{π}{3}, x = \frac{5 π}{3}

+1

度数

x = 6 0^{\circ}, x = 30 0^{\circ}

求解步骤

4 sin^{2} (x) = 5 - 4 cos (x), 0 \leq x < 2 π

两边减去

5 - 4 cos (x)

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

使用三角恒等式改写

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

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

化简

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

数字相乘:

4 \cdot 1 = 4

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

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

化简

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

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

对同类项分组

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

数字相加/相减:

- 5 + 4 = - 1

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

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

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

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

用替代法求解

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

令

: cos (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 = cos (x)

代回

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

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

cos (x) = \frac{1}{2}, 0 \leq x < 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

cos (x) = \frac{1}{2}, 0 \leq x < 2 π

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{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

在

0 \leq x < 2 π

范围内的解

x = \frac{π}{3}, x = \frac{5 π}{3}

合并所有解

x = \frac{π}{3}, x = \frac{5 π}{3}

作图

流行的例子

cos(θ)=sqrt(3/4)

cos (θ) = \frac{3}{4}

sqrt(2)sec(4x)-2=0

2 sec (4 x) - 2 = 0

solvefor A,y=-6cos(A)-2sin^2(A)

so l v e f or A, y = - 6 cos (A) - 2 sin^{2} (A)

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

sin^4(x)=cos^4(x)

sin^{4} (x) = cos^{4} (x)