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受欢迎的三角函数 >

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

解答

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

解答

x = 2 πn + \frac{13 π}{12}, x = 2 πn + \frac{17 π}{12}, x = 2 πn + \frac{π}{12}, x = 2 πn + \frac{5 π}{12}

+1

度数

x = 19 5^{\circ} + 36 0^{\circ} n, x = 25 5^{\circ} + 36 0^{\circ} n, x = 1 5^{\circ} + 36 0^{\circ} n, x = 7 5^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

\frac{3}{2}

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

化简

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

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

将项转换为分式:

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

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

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

分解

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

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

将

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

改写为

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

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

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

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

使用根式运算法则:

a = (a)^{2}

3 = (3)^{2}

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

使用指数法则:

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

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

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

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

使用平方差公式:

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

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

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

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

分别求解每个部分

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

2 (sin (x) + cos (x)) + 3 = 0 : x = 2 πn + \frac{13 π}{12}, x = 2 πn + \frac{17 π}{12}

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

使用三角恒等式改写

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

sin (x) + cos (x) = 2 sin (x + \frac{π}{4})

sin (x) + cos (x)

改写为

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

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{1}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (x) + sin (\frac{π}{4}) cos (x))

使用角和恒等式:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (x + \frac{π}{4})

= 3 + 22 sin (x + \frac{π}{4})

3 + 22 sin (x + \frac{π}{4}) = 0

使用根式运算法则:

a a = a

22 = 2

3 + 2 sin (x + \frac{π}{4}) = 0

将

3

到右边

3 + 2 sin (x + \frac{π}{4}) = 0

两边减去

3

3 + 2 sin (x + \frac{π}{4}) - 3 = 0 - 3

化简

2 sin (x + \frac{π}{4}) = - 3

2 sin (x + \frac{π}{4}) = - 3

两边除以

2

2 sin (x + \frac{π}{4}) = - 3

两边除以

2

\frac{2 sin ( x + \frac{π}{4} )}{2} = \frac{- 3}{2}

化简

sin (x + \frac{π}{4}) = - \frac{3}{2}

sin (x + \frac{π}{4}) = - \frac{3}{2}

sin (x + \frac{π}{4}) = - \frac{3}{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} = \frac{4 π}{3} + 2 πn, x + \frac{π}{4} = \frac{5 π}{3} + 2 πn

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

解

x + \frac{π}{4} = \frac{4 π}{3} + 2 πn : x = 2 πn + \frac{13 π}{12}

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

将

\frac{π}{4}

到右边

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

两边减去

\frac{π}{4}

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

化简

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

化简

x + \frac{π}{4} - \frac{π}{4} : x

x + \frac{π}{4} - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= x

化简

\frac{4 π}{3} + 2 πn - \frac{π}{4} : 2 πn + \frac{13 π}{12}

\frac{4 π}{3} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn - \frac{π}{4} + \frac{4 π}{3}

4, 3

的最小公倍数:

12

4, 3

最小公倍数 (LCM)

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

3

质因数分解:

3

3

3

是质数，因此无法因数分解

= 3

将每个因子乘以它在

4

或

3

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

12

对于

\frac{π}{4} :

将分母和分子乘以

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

对于

\frac{4 π}{3} :

将分母和分子乘以

4

\frac{4 π}{3} = \frac{4 π 4}{3 \cdot 4} = \frac{16 π}{12}

= - \frac{π 3}{12} + \frac{16 π}{12}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- π 3 + 16 π}{12}

同类项相加:

- 3 π + 16 π = 13 π

= 2 πn + \frac{13 π}{12}

x = 2 πn + \frac{13 π}{12}

x = 2 πn + \frac{13 π}{12}

x = 2 πn + \frac{13 π}{12}

解

x + \frac{π}{4} = \frac{5 π}{3} + 2 πn : x = 2 πn + \frac{17 π}{12}

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

将

\frac{π}{4}

到右边

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

两边减去

\frac{π}{4}

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

化简

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

化简

x + \frac{π}{4} - \frac{π}{4} : x

x + \frac{π}{4} - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= x

化简

\frac{5 π}{3} + 2 πn - \frac{π}{4} : 2 πn + \frac{17 π}{12}

\frac{5 π}{3} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn - \frac{π}{4} + \frac{5 π}{3}

4, 3

的最小公倍数:

12

4, 3

最小公倍数 (LCM)

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

3

质因数分解:

3

3

3

是质数，因此无法因数分解

= 3

将每个因子乘以它在

4

或

3

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

12

对于

\frac{π}{4} :

将分母和分子乘以

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

对于

\frac{5 π}{3} :

将分母和分子乘以

4

\frac{5 π}{3} = \frac{5 π 4}{3 \cdot 4} = \frac{20 π}{12}

= - \frac{π 3}{12} + \frac{20 π}{12}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- π 3 + 20 π}{12}

同类项相加:

- 3 π + 20 π = 17 π

= 2 πn + \frac{17 π}{12}

x = 2 πn + \frac{17 π}{12}

x = 2 πn + \frac{17 π}{12}

x = 2 πn + \frac{17 π}{12}

x = 2 πn + \frac{13 π}{12}, x = 2 πn + \frac{17 π}{12}

2 (sin (x) + cos (x)) - 3 = 0 : x = 2 πn + \frac{π}{12}, x = 2 πn + \frac{5 π}{12}

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

使用三角恒等式改写

2 (sin (x) + cos (x)) - 3

sin (x) + cos (x) = 2 sin (x + \frac{π}{4})

sin (x) + cos (x)

改写为

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

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{1}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (x) + sin (\frac{π}{4}) cos (x))

使用角和恒等式:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (x + \frac{π}{4})

= - 3 + 22 sin (x + \frac{π}{4})

- 3 + 22 sin (x + \frac{π}{4}) = 0

使用根式运算法则:

a a = a

22 = 2

- 3 + 2 sin (x + \frac{π}{4}) = 0

将

3

到右边

- 3 + 2 sin (x + \frac{π}{4}) = 0

两边加上

3

- 3 + 2 sin (x + \frac{π}{4}) + 3 = 0 + 3

化简

2 sin (x + \frac{π}{4}) = 3

2 sin (x + \frac{π}{4}) = 3

两边除以

2

2 sin (x + \frac{π}{4}) = 3

两边除以

2

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

化简

sin (x + \frac{π}{4}) = \frac{3}{2}

sin (x + \frac{π}{4}) = \frac{3}{2}

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

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

解

x + \frac{π}{4} = \frac{π}{3} + 2 πn : x = 2 πn + \frac{π}{12}

x + \frac{π}{4} = \frac{π}{3} + 2 πn

将

\frac{π}{4}

到右边

x + \frac{π}{4} = \frac{π}{3} + 2 πn

两边减去

\frac{π}{4}

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

化简

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

化简

x + \frac{π}{4} - \frac{π}{4} : x

x + \frac{π}{4} - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= x

化简

\frac{π}{3} + 2 πn - \frac{π}{4} : 2 πn + \frac{π}{12}

\frac{π}{3} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn + \frac{π}{3} - \frac{π}{4}

3, 4

的最小公倍数:

12

3, 4

最小公倍数 (LCM)

3

质因数分解:

3

3

3

是质数，因此无法因数分解

= 3

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

将每个因子乘以它在

3

或

4

中出现的最多次数

= 3 \cdot 2 \cdot 2

数字相乘:

3 \cdot 2 \cdot 2 = 12

= 12

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

12

对于

\frac{π}{3} :

将分母和分子乘以

4

\frac{π}{3} = \frac{π 4}{3 \cdot 4} = \frac{π 4}{12}

对于

\frac{π}{4} :

将分母和分子乘以

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

= \frac{π 4}{12} - \frac{π 3}{12}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{π 4 - π 3}{12}

同类项相加:

4 π - 3 π = π

= 2 πn + \frac{π}{12}

x = 2 πn + \frac{π}{12}

x = 2 πn + \frac{π}{12}

x = 2 πn + \frac{π}{12}

解

x + \frac{π}{4} = \frac{2 π}{3} + 2 πn : x = 2 πn + \frac{5 π}{12}

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

将

\frac{π}{4}

到右边

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

两边减去

\frac{π}{4}

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

化简

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

化简

x + \frac{π}{4} - \frac{π}{4} : x

x + \frac{π}{4} - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= x

化简

\frac{2 π}{3} + 2 πn - \frac{π}{4} : 2 πn + \frac{5 π}{12}

\frac{2 π}{3} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn - \frac{π}{4} + \frac{2 π}{3}

4, 3

的最小公倍数:

12

4, 3

最小公倍数 (LCM)

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

3

质因数分解:

3

3

3

是质数，因此无法因数分解

= 3

将每个因子乘以它在

4

或

3

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

12

对于

\frac{π}{4} :

将分母和分子乘以

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

对于

\frac{2 π}{3} :

将分母和分子乘以

4

\frac{2 π}{3} = \frac{2 π 4}{3 \cdot 4} = \frac{8 π}{12}

= - \frac{π 3}{12} + \frac{8 π}{12}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- π 3 + 8 π}{12}

同类项相加:

- 3 π + 8 π = 5 π

= 2 πn + \frac{5 π}{12}

x = 2 πn + \frac{5 π}{12}

x = 2 πn + \frac{5 π}{12}

x = 2 πn + \frac{5 π}{12}

x = 2 πn + \frac{π}{12}, x = 2 πn + \frac{5 π}{12}

合并所有解

x = 2 πn + \frac{13 π}{12}, x = 2 πn + \frac{17 π}{12}, x = 2 πn + \frac{π}{12}, x = 2 πn + \frac{5 π}{12}

作图

流行的例子

cos(x)=-1/(5sqrt(2))

cos (x) = - \frac{1}{5 2}

2cos^2(x)-sqrt(3cos(x))=0

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

sin(3x)=sin(5x)

sin (3 x) = sin (5 x)

2sin^2(3θ)+sin(3θ)-1=0

2 sin^{2} (3 θ) + sin (3 θ) - 1 = 0

4cos(θ)+sqrt(3)=2cos(θ)

4 cos (θ) + 3 = 2 cos (θ)