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受欢迎的三角函数 >

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

解答

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

解答

x \in R 无解

求解步骤

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

sin^{2} (x) + 2 (1 - sin^{2} (x)) : - sin^{2} (x) + 2

sin^{2} (x) + 2 (1 - sin^{2} (x))

乘开

2 (1 - sin^{2} (x)) : 2 - 2 sin^{2} (x)

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

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (x)

= 2 \cdot 1 - 2 sin^{2} (x)

数字相乘:

2 \cdot 1 = 2

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

= sin^{2} (x) + 2 - 2 sin^{2} (x)

化简

sin^{2} (x) + 2 - 2 sin^{2} (x) : - sin^{2} (x) + 2

sin^{2} (x) + 2 - 2 sin^{2} (x)

对同类项分组

= sin^{2} (x) - 2 sin^{2} (x) + 2

同类项相加:

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

= - sin^{2} (x) + 2

= - sin^{2} (x) + 2

= - sin^{2} (x) + 2

2 - sin^{2} (x) = 0

用替代法求解

2 - sin^{2} (x) = 0

令

: sin (x) = u

2 - u^{2} = 0

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

2 - u^{2} = 0

将

2

到右边

2 - u^{2} = 0

两边减去

2

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

化简

- u^{2} = - 2

- u^{2} = - 2

两边除以

- 1

- u^{2} = - 2

两边除以

- 1

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

化简

u^{2} = 2

u^{2} = 2

对于

x^{2} = f (a)

解为

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

u = 2, u = - 2

u = sin (x)

代回

sin (x) = 2, sin (x) = - 2

sin (x) = 2, sin (x) = - 2

sin (x) = 2 :

无解

sin (x) = 2

- 1 \leq sin (x) \leq 1

无解

sin (x) = - 2 :

无解

sin (x) = - 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

x \in R 无解

作图

流行的例子

sin(x+16)=cos(11x+2)

sin (x + 1 6^{\circ}) = cos (11 x + 2)

sin^4(x/3)+cos^4(x/3)= 5/8

sin^{4} (\frac{x}{3}) + cos^{4} (\frac{x}{3}) = \frac{5}{8}

- 2 sin (x) = 1

sec(x)cot(x)=-2

sec (x) cot (x) = - 2

- csc (x) = 1