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受欢迎的三角函数 >

3sin^2(θ)+2cos(2θ)=1,0<= θ<= 2pi

解答

3 sin^{2} (θ) + 2 cos (2 θ) = 1, 0 \leq θ \leq 2 π

解答

θ = \frac{π}{2}, θ = \frac{3 π}{2}

+1

度数

θ = 9 0^{\circ}, θ = 27 0^{\circ}

求解步骤

3 sin^{2} (θ) + 2 cos (2 θ) = 1, 0 \leq θ \leq 2 π

两边减去

1

3 sin^{2} (θ) + 2 cos (2 θ) - 1 = 0

使用三角恒等式改写

- 1 + 2 cos (2 θ) + 3 sin^{2} (θ)

使用倍角公式:

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

= - 1 + 2 (1 - 2 sin^{2} (θ)) + 3 sin^{2} (θ)

化简

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

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

乘开

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

2 (1 - 2 sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = 2 sin^{2} (θ)

= 2 \cdot 1 - 2 \cdot 2 sin^{2} (θ)

化简

2 \cdot 1 - 2 \cdot 2 sin^{2} (θ) : 2 - 4 sin^{2} (θ)

2 \cdot 1 - 2 \cdot 2 sin^{2} (θ)

数字相乘:

2 \cdot 1 = 2

= 2 - 2 \cdot 2 sin^{2} (θ)

数字相乘:

2 \cdot 2 = 4

= 2 - 4 sin^{2} (θ)

= 2 - 4 sin^{2} (θ)

= - 1 + 2 - 4 sin^{2} (θ) + 3 sin^{2} (θ)

化简

- 1 + 2 - 4 sin^{2} (θ) + 3 sin^{2} (θ) : - sin^{2} (θ) + 1

- 1 + 2 - 4 sin^{2} (θ) + 3 sin^{2} (θ)

同类项相加:

- 4 sin^{2} (θ) + 3 sin^{2} (θ) = - sin^{2} (θ)

= - 1 + 2 - sin^{2} (θ)

数字相加/相减:

- 1 + 2 = 1

= - sin^{2} (θ) + 1

= - sin^{2} (θ) + 1

= - sin^{2} (θ) + 1

1 - sin^{2} (θ) = 0

用替代法求解

1 - sin^{2} (θ) = 0

令

: sin (θ) = u

1 - u^{2} = 0

1 - u^{2} = 0 : u = 1, u = - 1

1 - u^{2} = 0

将

1

到右边

1 - u^{2} = 0

两边减去

1

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

化简

- u^{2} = - 1

- u^{2} = - 1

两边除以

- 1

- u^{2} = - 1

两边除以

- 1

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

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

u = sin (θ)

代回

sin (θ) = 1, sin (θ) = - 1

sin (θ) = 1, sin (θ) = - 1

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

sin (θ) = 1, 0 \leq θ \leq 2 π

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{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

在

0 \leq θ \leq 2 π

范围内的解

θ = \frac{π}{2}

sin (θ) = - 1, 0 \leq θ \leq 2 π : θ = \frac{3 π}{2}

sin (θ) = - 1, 0 \leq θ \leq 2 π

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

在

0 \leq θ \leq 2 π

范围内的解

θ = \frac{3 π}{2}

合并所有解

θ = \frac{π}{2}, θ = \frac{3 π}{2}

作图

流行的例子

2cos^2(x)+3sin(-x)-3=0

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

cos (π x) = 1

solvefor x,pi+3arccos(x+1)=0

so l v e f or x, π + 3 arccos (x + 1) = 0

cos(3x)=-1,0<= x<= 2pi

cos (3 x) = - 1, 0 \leq x \leq 2 π

sin(x)=-sqrt(3)cos(x)

sin (x) = - 3 cos (x)