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受欢迎的三角函数 >

sin^2(θ)=2cos(θ)+1

解答

sin^{2} (θ) = 2 cos (θ) + 1

解答

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

+1

度数

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

求解步骤

sin^{2} (θ) = 2 cos (θ) + 1

两边减去

2 cos (θ) + 1

sin^{2} (θ) - 2 cos (θ) - 1 = 0

使用三角恒等式改写

- 1 + sin^{2} (θ) - 2 cos (θ)

使用毕达哥拉斯恒等式:

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

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

= - 2 cos (θ) - cos^{2} (θ)

- cos^{2} (θ) - 2 cos (θ) = 0

用替代法求解

- cos^{2} (θ) - 2 cos (θ) = 0

令

: cos (θ) = u

- u^{2} - 2 u = 0

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

- u^{2} - 2 u = 0

使用求根公式求解

- u^{2} - 2 u = 0

二次方程求根公式:

若

a = - 1, b = - 2, c = 0

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

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

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

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

使用法则

- (- a) = a

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

将解分隔开

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

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

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

去除括号:

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

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

数字相加:

2 + 2 = 4

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

数字相乘:

2 \cdot 1 = 2

= \frac{4}{- 2}

使用分式法则:

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

= - \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= - 2

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

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

去除括号:

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

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

数字相减:

2 - 2 = 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

二次方程组的解是:

u = - 2, u = 0

u = cos (θ)

代回

cos (θ) = - 2, cos (θ) = 0

cos (θ) = - 2, cos (θ) = 0

cos (θ) = - 2 :

无解

cos (θ) = - 2

- 1 \leq cos (x) \leq 1

无解

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

cos (θ) = 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}

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

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

合并所有解

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

作图

流行的例子

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(cos (θ)) (cos (θ) + 1) = 0

5-tan^2(x)=4,0<= x<= 360

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