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受欢迎的三角函数 >

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

解答

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

解答

x = 2 πn, x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

1

2 sin^{3} (x) + cos^{2} (x) - 1 = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

用替代法求解

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

令

: sin (x) = u

- u^{2} + 2 u^{3} = 0

- u^{2} + 2 u^{3} = 0 : u = 0, u = \frac{1}{2}

- u^{2} + 2 u^{3} = 0

因式分解

- u^{2} + 2 u^{3} : u^{2} (2 u - 1)

- u^{2} + 2 u^{3}

使用指数法则:

a^{b + c} = a^{b} a^{c}

u^{3} = u u^{2}

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

因式分解出通项

u^{2}

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

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

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u = 0 or 2 u - 1 = 0

解

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

2 u - 1 = 0

将

1

到右边

2 u - 1 = 0

两边加上

1

2 u - 1 + 1 = 0 + 1

化简

2 u = 1

2 u = 1

两边除以

2

2 u = 1

两边除以

2

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

化简

u = \frac{1}{2}

u = \frac{1}{2}

解为

u = 0, u = \frac{1}{2}

u = sin (x)

代回

sin (x) = 0, sin (x) = \frac{1}{2}

sin (x) = 0, sin (x) = \frac{1}{2}

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

sin (x) = \frac{1}{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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

合并所有解

x = 2 πn, x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

作图

流行的例子

2tan(x)=tan(x)tan(x)

2 tan (x) = tan (x) tan (x)

(2tan(x))/(1-tan^2(x))=sqrt(3)

\frac{2 tan ( x )}{1 - tan ^{2} ( x )} = 3

tan (x) = \frac{59}{56}

cos(θ)= 5/(5sqrt(2))

cos (θ) = \frac{5}{5 2}

arccos(x)=arcsin(9/41)

arccos (x) = arcsin (\frac{9}{41})