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受欢迎的三角函数 >

1=3cos(2θ)

解答

1 = 3 cos (2 θ)

解答

θ = \frac{1.23095 \dots}{2} + πn, θ = π - \frac{1.23095 \dots}{2} + πn

+1

度数

θ = 35.26438 \dots^{\circ} + 18 0^{\circ} n, θ = 144.73561 \dots^{\circ} + 18 0^{\circ} n

求解步骤

1 = 3 cos (2 θ)

交换两边

3 cos (2 θ) = 1

两边除以

3

3 cos (2 θ) = 1

两边除以

3

\frac{3 cos ( 2 θ )}{3} = \frac{1}{3}

化简

cos (2 θ) = \frac{1}{3}

cos (2 θ) = \frac{1}{3}

使用反三角函数性质

cos (2 θ) = \frac{1}{3}

cos (2 θ) = \frac{1}{3}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

2 θ = arccos (\frac{1}{3}) + 2 πn, 2 θ = 2 π - arccos (\frac{1}{3}) + 2 πn

2 θ = arccos (\frac{1}{3}) + 2 πn, 2 θ = 2 π - arccos (\frac{1}{3}) + 2 πn

解

2 θ = arccos (\frac{1}{3}) + 2 πn : θ = \frac{arccos ( \frac{1}{3} )}{2} + πn

2 θ = arccos (\frac{1}{3}) + 2 πn

两边除以

2

2 θ = arccos (\frac{1}{3}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{arccos ( \frac{1}{3} )}{2} + \frac{2 πn}{2}

化简

θ = \frac{arccos ( \frac{1}{3} )}{2} + πn

θ = \frac{arccos ( \frac{1}{3} )}{2} + πn

解

2 θ = 2 π - arccos (\frac{1}{3}) + 2 πn : θ = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

2 θ = 2 π - arccos (\frac{1}{3}) + 2 πn

两边除以

2

2 θ = 2 π - arccos (\frac{1}{3}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{1}{3} )}{2} + \frac{2 πn}{2}

化简

θ = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

θ = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

θ = \frac{arccos ( \frac{1}{3} )}{2} + πn, θ = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

以小数形式表示解

θ = \frac{1.23095 \dots}{2} + πn, θ = π - \frac{1.23095 \dots}{2} + πn

作图

流行的例子

4sin(x)=4sin(2x)

4 sin (x) = 4 sin (2 x)

csc(x)-sin(x)=cos(x)cot(3x-50)

csc (x) - sin (x) = cos (x) cot (3 x - 50)

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

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

tan^2(t)-sec^2(t)=1

tan^{2} (t) - sec^{2} (t) = 1

solvefor x,z=sin(2x+3y)

so l v e f or x, z = sin (2 x + 3 y)