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受欢迎的三角函数 >

-2sin(θ)+cos(2θ)=1

解答

- 2 sin (θ) + cos (2 θ) = 1

解答

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

+1

度数

θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

- 2 sin (θ) + cos (2 θ) = 1

两边减去

1

- 2 sin (θ) + cos (2 θ) - 1 = 0

使用三角恒等式改写

- 1 + cos (2 θ) - 2 sin (θ)

使用倍角公式:

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

= - 1 + 1 - 2 sin^{2} (θ) - 2 sin (θ)

化简

= - 2 sin^{2} (θ) - 2 sin (θ)

- 2 sin (θ) - 2 sin^{2} (θ) = 0

用替代法求解

- 2 sin (θ) - 2 sin^{2} (θ) = 0

令

: sin (θ) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 2, b = - 2, c = 0

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

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

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

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

使用法则

- (- a) = a

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

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 2)^{2} = 2^{2}

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

使用法则

0 \cdot a = 0

= 2^{2} + 0

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

= 2^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 2

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

将解分隔开

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

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

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

去除括号:

(- a) = - a, - (- a) = a

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

数字相加:

2 + 2 = 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{- 4}

使用分式法则:

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

= - \frac{4}{4}

使用法则

\frac{a}{a} = 1

= - 1

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

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

去除括号:

(- a) = - a, - (- a) = a

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

数字相减:

2 - 2 = 0

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

数字相乘:

2 \cdot 2 = 4

= \frac{0}{- 4}

使用分式法则:

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

= - \frac{0}{4}

使用法则

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

= - 0

= 0

二次方程组的解是:

u = - 1, u = 0

u = sin (θ)

代回

sin (θ) = - 1, sin (θ) = 0

sin (θ) = - 1, sin (θ) = 0

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

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

sin (θ) = 0

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 = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

合并所有解

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

作图

流行的例子

cos(2x+60)=sin(3x)

cos (2 x + 60) = sin (3 x)

csc (θ) = - \frac{3}{2}

4 cos (θ) = 0

(sin(160))/5 =(sin(c))/4

\frac{sin ( 16 0 ^{\circ} )}{5} = \frac{sin ( c )}{4}

sin(x)+sqrt(2)=0

sin (x) + 2 = 0