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受欢迎的三角函数 >

4cos(x)=-sin^2(x)+1

解答

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

解答

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

+1

度数

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

求解步骤

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

两边减去

- sin^{2} (x) + 1

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

用替代法求解

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

令

: cos (x) = u

- u^{2} + 4 u = 0

- u^{2} + 4 u = 0 : u = 0, u = 4

- u^{2} + 4 u = 0

使用求根公式求解

- u^{2} + 4 u = 0

二次方程求根公式:

若

a = - 1, b = 4, c = 0

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

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

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

4^{2} - 4 (- 1) \cdot 0

使用法则

- (- a) = a

= 4^{2} + 4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 4^{2} + 0

4^{2} + 0 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 4

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

将解分隔开

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

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

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

去除括号:

(- a) = - a

= \frac{- 4 + 4}{- 2 \cdot 1}

数字相加/相减:

- 4 + 4 = 0

= \frac{0}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{- 2}

使用分式法则:

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

= - \frac{0}{2}

使用法则

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

= - 0

= 0

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

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

去除括号:

(- a) = - a

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

数字相减:

- 4 - 4 = - 8

= \frac{- 8}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 8}{- 2}

使用分式法则:

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

= \frac{8}{2}

数字相除:

\frac{8}{2} = 4

= 4

二次方程组的解是:

u = 0, u = 4

u = cos (x)

代回

cos (x) = 0, cos (x) = 4

cos (x) = 0, cos (x) = 4

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

cos (x) = 4 :

无解

cos (x) = 4

- 1 \leq cos (x) \leq 1

无解

合并所有解

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

作图

流行的例子

sin (\frac{x}{4}) = 0

sin((4pi)/(11))=sin(θ)*sin((7pi)/(11))

sin (\frac{4 π}{11}) = sin (θ) \cdot sin (\frac{7 π}{11})

sin(2x)=2tan(2x)

sin (2 x) = 2 tan (2 x)

-2cos(2θ)+12=4sin(θ)+9

- 2 cos (2 θ) + 12 = 4 sin (θ) + 9

sec^2(x)+sec(x)-2=0

sec^{2} (x) + sec (x) - 2 = 0