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受欢迎的三角函数 >

cos(2x)+7sin(x)-4=0

解答

cos (2 x) + 7 sin (x) - 4 = 0

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

cos (2 x) + 7 sin (x) - 4 = 0

使用三角恒等式改写

- 4 + cos (2 x) + 7 sin (x)

使用倍角公式:

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

= - 4 + 1 - 2 sin^{2} (x) + 7 sin (x)

化简

= 7 sin (x) - 2 sin^{2} (x) - 3

- 3 - 2 sin^{2} (x) + 7 sin (x) = 0

用替代法求解

- 3 - 2 sin^{2} (x) + 7 sin (x) = 0

令

: sin (x) = u

- 3 - 2 u^{2} + 7 u = 0

- 3 - 2 u^{2} + 7 u = 0 : u = \frac{1}{2}, u = 3

- 3 - 2 u^{2} + 7 u = 0

改写成标准形式

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

- 2 u^{2} + 7 u - 3 = 0

使用求根公式求解

- 2 u^{2} + 7 u - 3 = 0

二次方程求根公式:

若

a = - 2, b = 7, c = - 3

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 2 ) ( - 3 )}{2 ( - 2 )}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 2 ) ( - 3 )}{2 ( - 2 )}

7^{2} - 4 (- 2) (- 3) = 5

7^{2} - 4 (- 2) (- 3)

使用法则

- (- a) = a

= 7^{2} - 4 \cdot 2 \cdot 3

数字相乘:

4 \cdot 2 \cdot 3 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

数字相减:

49 - 24 = 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{- 7 \pm 5}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 7 + 5}{2 ( - 2 )}, u_{2} = \frac{- 7 - 5}{2 ( - 2 )}

u = \frac{- 7 + 5}{2 ( - 2 )} : \frac{1}{2}

\frac{- 7 + 5}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 7 + 5}{- 2 \cdot 2}

数字相加/相减:

- 7 + 5 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

使用分式法则:

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

= \frac{2}{4}

约分:

2

= \frac{1}{2}

u = \frac{- 7 - 5}{2 ( - 2 )} : 3

\frac{- 7 - 5}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 7 - 5}{- 2 \cdot 2}

数字相减:

- 7 - 5 = - 12

= \frac{- 12}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 12}{- 4}

使用分式法则:

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

= \frac{12}{4}

数字相除:

\frac{12}{4} = 3

= 3

二次方程组的解是:

u = \frac{1}{2}, u = 3

u = sin (x)

代回

sin (x) = \frac{1}{2}, sin (x) = 3

sin (x) = \frac{1}{2}, sin (x) = 3

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

sin (x) = 3 :

无解

sin (x) = 3

- 1 \leq sin (x) \leq 1

无解

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(θ/8)+sqrt(3)=0

tan (\frac{θ}{8}) + 3 = 0

sin(θ/2)-1=0,0<= θ<= 2pi

sin (\frac{θ}{2}) - 1 = 0, 0 \leq θ \leq 2 π

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

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

(3+sin^2(θ))/(cos(θ)-2)=3cos(θ)

\frac{3 + sin ^{2} ( θ )}{cos ( θ ) - 2} = 3 cos (θ)

cos (a) = 0