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受欢迎的三角函数 >

9+9sin(θ)=6cos^2(θ)

解答

9 + 9 sin (θ) = 6 cos^{2} (θ)

解答

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

+1

度数

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

9 + 9 sin (θ) = 6 cos^{2} (θ)

两边减去

6 cos^{2} (θ)

9 + 9 sin (θ) - 6 cos^{2} (θ) = 0

使用三角恒等式改写

9 - 6 cos^{2} (θ) + 9 sin (θ)

使用毕达哥拉斯恒等式:

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

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

= 9 - 6 (1 - sin^{2} (θ)) + 9 sin (θ)

化简

9 - 6 (1 - sin^{2} (θ)) + 9 sin (θ) : 6 sin^{2} (θ) + 9 sin (θ) + 3

9 - 6 (1 - sin^{2} (θ)) + 9 sin (θ)

乘开

- 6 (1 - sin^{2} (θ)) : - 6 + 6 sin^{2} (θ)

- 6 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 6, b = 1, c = sin^{2} (θ)

= - 6 \cdot 1 - (- 6) sin^{2} (θ)

使用加减运算法则

- (- a) = a

= - 6 \cdot 1 + 6 sin^{2} (θ)

数字相乘:

6 \cdot 1 = 6

= - 6 + 6 sin^{2} (θ)

= 9 - 6 + 6 sin^{2} (θ) + 9 sin (θ)

数字相减:

9 - 6 = 3

= 6 sin^{2} (θ) + 9 sin (θ) + 3

= 6 sin^{2} (θ) + 9 sin (θ) + 3

3 + 6 sin^{2} (θ) + 9 sin (θ) = 0

用替代法求解

3 + 6 sin^{2} (θ) + 9 sin (θ) = 0

令

: sin (θ) = u

3 + 6 u^{2} + 9 u = 0

3 + 6 u^{2} + 9 u = 0 : u = - \frac{1}{2}, u = - 1

3 + 6 u^{2} + 9 u = 0

改写成标准形式

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

6 u^{2} + 9 u + 3 = 0

使用求根公式求解

6 u^{2} + 9 u + 3 = 0

二次方程求根公式:

若

a = 6, b = 9, c = 3

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 6 \cdot 3}{2 \cdot 6}

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 6 \cdot 3}{2 \cdot 6}

9^{2} - 4 \cdot 6 \cdot 3 = 3

9^{2} - 4 \cdot 6 \cdot 3

数字相乘:

4 \cdot 6 \cdot 3 = 72

= 9^{2} - 72

9^{2} = 81

= 81 - 72

数字相减:

81 - 72 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- 9 \pm 3}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- 9 + 3}{2 \cdot 6}, u_{2} = \frac{- 9 - 3}{2 \cdot 6}

u = \frac{- 9 + 3}{2 \cdot 6} : - \frac{1}{2}

\frac{- 9 + 3}{2 \cdot 6}

数字相加/相减:

- 9 + 3 = - 6

= \frac{- 6}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 6}{12}

使用分式法则:

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

= - \frac{6}{12}

约分:

6

= - \frac{1}{2}

u = \frac{- 9 - 3}{2 \cdot 6} : - 1

\frac{- 9 - 3}{2 \cdot 6}

数字相减:

- 9 - 3 = - 12

= \frac{- 12}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 12}{12}

使用分式法则:

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

= - \frac{12}{12}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

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

u = sin (θ)

代回

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

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

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - \frac{1}{2}

sin (θ) = - \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}

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

sin (θ) = - 1

sin (θ) = - 1

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}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

合并所有解

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

作图

流行的例子

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

sec (t) = 3

tan (θ) = - \frac{12}{5}

solvefor x,sin(2x)=sin(x)

so l v e f or x, sin (2 x) = sin (x)

-3cos(2θ)+7cos(θ)-10=-5

- 3 cos (2 θ) + 7 cos (θ) - 10 = - 5