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受欢迎的三角函数 >

sin^2(x)=2cos^4(x)

解答

sin^{2} (x) = 2 cos^{4} (x)

解答

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

+1

度数

x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n

求解步骤

sin^{2} (x) = 2 cos^{4} (x)

两边减去

2 cos^{4} (x)

sin^{2} (x) - 2 cos^{4} (x) = 0

分解

sin^{2} (x) - 2 cos^{4} (x) : (sin (x) + 2 cos^{2} (x)) (sin (x) - 2 cos^{2} (x))

sin^{2} (x) - 2 cos^{4} (x)

将

sin^{2} (x) - 2 cos^{4} (x)

改写为

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

sin^{2} (x) - 2 cos^{4} (x)

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

= sin^{2} (x) - (2)^{2} cos^{4} (x)

使用指数法则:

a^{b c} = (a^{b})^{c}

cos^{4} (x) = (cos^{2} (x))^{2}

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

使用指数法则:

a^{m} b^{m} = (ab)^{m}

(2)^{2} (cos^{2} (x))^{2} = (2 cos^{2} (x))^{2}

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

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

使用平方差公式:

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

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

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

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

分别求解每个部分

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

sin (x) + 2 cos^{2} (x) = 0 : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

使用三角恒等式改写

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

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

sin (x) + (1 - sin^{2} (x)) 2 = 0

用替代法求解

sin (x) + (1 - sin^{2} (x)) 2 = 0

令

: sin (x) = u

u + (1 - u^{2}) 2 = 0

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

u + (1 - u^{2}) 2 = 0

展开

u + (1 - u^{2}) 2 : u + 2 - 2 u^{2}

u + (1 - u^{2}) 2

= u + 2 (1 - u^{2})

乘开

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

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

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

乘以:

1 \cdot 2 = 2

= 2 - 2 u^{2}

= u + 2 - 2 u^{2}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

1^{2} - 4 (- 2) 2 = 3

1^{2} - 4 (- 2) 2

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 42 (- 2)

使用法则

- (- a) = a

= 1 + 422

422 = 8

422

使用根式运算法则:

a a = a

22 = 2

= 4 \cdot 2

数字相乘:

4 \cdot 2 = 8

= 8

= 1 + 8

数字相加:

1 + 8 = 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{- 1 \pm 3}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 1 + 3}{2 ( - 2 )}, u_{2} = \frac{- 1 - 3}{2 ( - 2 )}

u = \frac{- 1 + 3}{2 ( - 2 )} : - \frac{2}{2}

\frac{- 1 + 3}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 1 + 3}{- 2 2}

数字相加/相减:

- 1 + 3 = 2

= \frac{2}{- 2 2}

使用分式法则:

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

= - \frac{2}{2 2}

数字相除:

\frac{2}{2} = 1

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

u = \frac{- 1 - 3}{2 ( - 2 )} : 2

\frac{- 1 - 3}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 1 - 3}{- 2 2}

数字相减:

- 1 - 3 = - 4

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

使用分式法则:

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

= \frac{4}{2 2}

数字相除:

\frac{4}{2} = 2

= \frac{2}{2}

使用根式运算法则:

n a = a^{\frac{1}{n}}

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

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

使用指数法则:

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

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

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

数字相减:

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

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

使用根式运算法则:

a^{\frac{1}{n}} = n a

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

= 2

二次方程组的解是:

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

u = sin (x)

代回

sin (x) = - \frac{2}{2}, sin (x) = 2

sin (x) = - \frac{2}{2}, sin (x) = 2

sin (x) = - \frac{2}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

sin (x) = - \frac{2}{2}

sin (x) = - \frac{2}{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}

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

sin (x) = 2 :

无解

sin (x) = 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

sin (x) - 2 cos^{2} (x) = 0 : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

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

使用三角恒等式改写

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

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

sin (x) - (1 - sin^{2} (x)) 2 = 0

用替代法求解

sin (x) - (1 - sin^{2} (x)) 2 = 0

令

: sin (x) = u

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

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

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

展开

u - (1 - u^{2}) 2 : u - 2 + 2 u^{2}

u - (1 - u^{2}) 2

= u - 2 (1 - u^{2})

乘开

- 2 (1 - u^{2}) : - 2 + 2 u^{2}

- 2 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = - 2, b = 1, c = u^{2}

= - 2 \cdot 1 - (- 2) u^{2}

使用加减运算法则

- (- a) = a

= - 1 \cdot 2 + 2 u^{2}

乘以:

1 \cdot 2 = 2

= - 2 + 2 u^{2}

= u - 2 + 2 u^{2}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

1^{2} - 42 (- 2) = 3

1^{2} - 42 (- 2)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 42 (- 2)

使用法则

- (- a) = a

= 1 + 422

422 = 8

422

使用根式运算法则:

a a = a

22 = 2

= 4 \cdot 2

数字相乘:

4 \cdot 2 = 8

= 8

= 1 + 8

数字相加:

1 + 8 = 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{- 1 \pm 3}{2 2}

将解分隔开

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

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

\frac{- 1 + 3}{2 2}

数字相加/相减:

- 1 + 3 = 2

= \frac{2}{2 2}

数字相除:

\frac{2}{2} = 1

= \frac{1}{2}

\frac{1}{2}

有理化:

\frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

= \frac{2}{2}

u = \frac{- 1 - 3}{2 2} : - 2

\frac{- 1 - 3}{2 2}

数字相减:

- 1 - 3 = - 4

= \frac{- 4}{2 2}

使用分式法则:

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

= - \frac{4}{2 2}

数字相除:

\frac{4}{2} = 2

= \frac{2}{2}

使用根式运算法则:

n a = a^{\frac{1}{n}}

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

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

使用指数法则:

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

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

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

数字相减:

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

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

使用根式运算法则:

a^{\frac{1}{n}} = n a

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

= - 2

二次方程组的解是:

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

u = sin (x)

代回

sin (x) = \frac{2}{2}, sin (x) = - 2

sin (x) = \frac{2}{2}, sin (x) = - 2

sin (x) = \frac{2}{2} : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

sin (x) = \frac{2}{2}

sin (x) = \frac{2}{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}

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

sin (x) = - 2 :

无解

sin (x) = - 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

合并所有解

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

作图

流行的例子

sin^3(x)+cos^3(x)=(1-1)/(2sin^2(x))

sin^{3} (x) + cos^{3} (x) = \frac{1 - 1}{2 sin ^{2} ( x )}

solvefor i,xsin^2(x)=cos^2(x)

so l v e f or i, x sin^{2} (x) = cos^{2} (x)

solvefor c,sin^2(x)+12=12(sin(x)-cos(x))

so l v e f or c, sin^{2} (x) + 12 = 12 (sin (x) - cos (x))

solvefor c,sin(x)-5cos(x)=sin^2(x)+5cos^2(x)

so l v e f or c, sin (x) - 5 cos (x) = sin^{2} (x) + 5 cos^{2} (x)

solvefor i,sin(x)+sin^2(x)+cos^3(x)=0

so l v e f or i, sin (x) + sin^{2} (x) + cos^{3} (x) = 0