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2asin(x)+a=0

解答

2 a sin (x) + a = 0

解答

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

+1

度数

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

求解步骤

2 a sin (x) + a = 0

将

a

到右边

2 a sin (x) + a = 0

两边减去

a

2 a sin (x) + a - a = 0 - a

化简

2 a sin (x) = - a

2 a sin (x) = - a

两边除以

2 a; a \neq = 0

2 a sin (x) = - a

两边除以

2 a; a \neq = 0

\frac{2 a sin ( x )}{2 a} = \frac{- a}{2 a}; a \neq = 0

化简

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

化简

\frac{2 a sin ( x )}{2 a} : sin (x)

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

数字相除:

\frac{2}{2} = 1

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

约分:

a

= sin (x)

化简

\frac{- a}{2 a} : - \frac{1}{2}

\frac{- a}{2 a}

使用分式法则:

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

= - \frac{a}{2 a}

约分:

a

= - \frac{1}{2}

sin (x) = - \frac{1}{2}; a \neq = 0

sin (x) = - \frac{1}{2}; a \neq = 0

sin (x) = - \frac{1}{2}; a \neq = 0

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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

流行的例子

tan (θ) = (\frac{2}{3})

tan (θ) = 8.1443

4sin(x)=5cos(x)

4 sin (x) = 5 cos (x)

tan (θ) = \frac{26}{41}

cot(θ)+2csc(θ)=4,0<= θ<= 360

cot (θ) + 2 csc (θ) = 4, 0^{\circ} \leq θ \leq 36 0^{\circ}