zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

cos(x-15^0)=((sqrt(2)))/2

解答

cos (x - 1 5^{0}) = \frac{( 2 )}{2}

解答

x = \frac{π}{4} + 2 πn + 1, x = \frac{7 π}{4} + 2 πn + 1

+1

度数

x = 102.29577 \dots^{\circ} + 36 0^{\circ} n, x = 372.29577 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cos (x - 1 5^{0}) = \frac{( 2 )}{2}

cos (x - 1 5^{0}) = \frac{2}{2}

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}

x - 1 5^{0} = \frac{π}{4} + 2 πn, x - 1 5^{0} = \frac{7 π}{4} + 2 πn

x - 1 5^{0} = \frac{π}{4} + 2 πn, x - 1 5^{0} = \frac{7 π}{4} + 2 πn

解

x - 1 5^{0} = \frac{π}{4} + 2 πn : x = \frac{π}{4} + 2 πn + 1

x - 1 5^{0} = \frac{π}{4} + 2 πn

使用法则

a^{0} = 1, a \neq = 0

1 5^{0} = 1

x - 1 = \frac{π}{4} + 2 πn

将

1

到右边

x - 1 = \frac{π}{4} + 2 πn

两边加上

1

x - 1 + 1 = \frac{π}{4} + 2 πn + 1

化简

x = \frac{π}{4} + 2 πn + 1

x = \frac{π}{4} + 2 πn + 1

解

x - 1 5^{0} = \frac{7 π}{4} + 2 πn : x = \frac{7 π}{4} + 2 πn + 1

x - 1 5^{0} = \frac{7 π}{4} + 2 πn

使用法则

a^{0} = 1, a \neq = 0

1 5^{0} = 1

x - 1 = \frac{7 π}{4} + 2 πn

将

1

到右边

x - 1 = \frac{7 π}{4} + 2 πn

两边加上

1

x - 1 + 1 = \frac{7 π}{4} + 2 πn + 1

化简

x = \frac{7 π}{4} + 2 πn + 1

x = \frac{7 π}{4} + 2 πn + 1

x = \frac{π}{4} + 2 πn + 1, x = \frac{7 π}{4} + 2 πn + 1

作图

流行的例子

sin(x)=1-cos^x(x)

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

4-7sin(x)=cos^2(x)

4 - 7 sin (x) = cos^{2} (x)

(d^2-2d+3)=sin(x)

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

\frac{5}{tan ( x )} = - 2

tan^{22}(x)=sec^2(x)-1

tan^{22} (x) = sec^{2} (x) - 1