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受欢迎的三角函数 >

solvefor t,cos(wt+d)=0

解答

求解

t, cos (wt + d) = 0

解答

t = \frac{π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}, t = \frac{3 π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}; w \neq = 0

求解步骤

cos (wt + d) = 0

cos (wt + d) = 0

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}

wt + d = \frac{π}{2} + 2 πn, wt + d = \frac{3 π}{2} + 2 πn

wt + d = \frac{π}{2} + 2 πn, wt + d = \frac{3 π}{2} + 2 πn

解

wt + d = \frac{π}{2} + 2 πn : t = \frac{π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}; w \neq = 0

wt + d = \frac{π}{2} + 2 πn

将

d

到右边

wt + d = \frac{π}{2} + 2 πn

两边减去

d

wt + d - d = \frac{π}{2} + 2 πn - d

化简

wt = \frac{π}{2} + 2 πn - d

wt = \frac{π}{2} + 2 πn - d

两边除以

w; w \neq = 0

wt = \frac{π}{2} + 2 πn - d

两边除以

w; w \neq = 0

\frac{wt}{w} = \frac{\frac{π}{2}}{w} + \frac{2 πn}{w} - \frac{d}{w}; w \neq = 0

化简

\frac{wt}{w} = \frac{\frac{π}{2}}{w} + \frac{2 πn}{w} - \frac{d}{w}

化简

\frac{wt}{w} : t

\frac{wt}{w}

约分:

w

= t

化简

\frac{\frac{π}{2}}{w} + \frac{2 πn}{w} - \frac{d}{w} : \frac{π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}

\frac{\frac{π}{2}}{w} + \frac{2 πn}{w} - \frac{d}{w}

使用分式法则:

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

= \frac{π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}

t = \frac{π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}; w \neq = 0

t = \frac{π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}; w \neq = 0

t = \frac{π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}; w \neq = 0

解

wt + d = \frac{3 π}{2} + 2 πn : t = \frac{3 π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}; w \neq = 0

wt + d = \frac{3 π}{2} + 2 πn

将

d

到右边

wt + d = \frac{3 π}{2} + 2 πn

两边减去

d

wt + d - d = \frac{3 π}{2} + 2 πn - d

化简

wt = \frac{3 π}{2} + 2 πn - d

wt = \frac{3 π}{2} + 2 πn - d

两边除以

w; w \neq = 0

wt = \frac{3 π}{2} + 2 πn - d

两边除以

w; w \neq = 0

\frac{wt}{w} = \frac{\frac{3 π}{2}}{w} + \frac{2 πn}{w} - \frac{d}{w}; w \neq = 0

化简

\frac{wt}{w} = \frac{\frac{3 π}{2}}{w} + \frac{2 πn}{w} - \frac{d}{w}

化简

\frac{wt}{w} : t

\frac{wt}{w}

约分:

w

= t

化简

\frac{\frac{3 π}{2}}{w} + \frac{2 πn}{w} - \frac{d}{w} : \frac{3 π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}

\frac{\frac{3 π}{2}}{w} + \frac{2 πn}{w} - \frac{d}{w}

使用分式法则:

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

= \frac{3 π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}

t = \frac{3 π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}; w \neq = 0

t = \frac{3 π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}; w \neq = 0

t = \frac{3 π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}; w \neq = 0

t = \frac{π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}, t = \frac{3 π}{2 w} + \frac{2 πn}{w} - \frac{d}{w}; w \neq = 0

作图

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