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受欢迎的三角函数 >

2sin(2x)=cos(2x+30)

解答

2 sin (2 x) = cos (2 x + 3 0^{\circ})

解答

x = \frac{0.33347 \dots}{2} + \frac{18 0 ^{\circ} n}{2}

+1

弧度

x = \frac{0.33347 \dots}{2} + \frac{π}{2} n

求解步骤

2 sin (2 x) = cos (2 x + 3 0^{\circ})

使用三角恒等式改写

2 sin (2 x) = cos (2 x + 3 0^{\circ})

使用三角恒等式改写

cos (2 x + 3 0^{\circ})

使用角和恒等式:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

= cos (2 x) cos (3 0^{\circ}) - sin (2 x) sin (3 0^{\circ})

化简

cos (2 x) cos (3 0^{\circ}) - sin (2 x) sin (3 0^{\circ}) : \frac{3}{2} cos (2 x) - \frac{1}{2} sin (2 x)

cos (2 x) cos (3 0^{\circ}) - sin (2 x) sin (3 0^{\circ})

化简

cos (3 0^{\circ}) : \frac{3}{2}

cos (3 0^{\circ})

使用以下普通恒等式:

cos (3 0^{\circ}) = \frac{3}{2}

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

= \frac{3}{2} cos (2 x) - sin (3 0^{\circ}) sin (2 x)

化简

sin (3 0^{\circ}) : \frac{1}{2}

sin (3 0^{\circ})

使用以下普通恒等式:

sin (3 0^{\circ}) = \frac{1}{2}

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

= \frac{3}{2} cos (2 x) - \frac{1}{2} sin (2 x)

= \frac{3}{2} cos (2 x) - \frac{1}{2} sin (2 x)

2 sin (2 x) = \frac{3}{2} cos (2 x) - \frac{1}{2} sin (2 x)

2 sin (2 x) = \frac{3}{2} cos (2 x) - \frac{1}{2} sin (2 x)

两边减去

\frac{3}{2} cos (2 x) - \frac{1}{2} sin (2 x)

\frac{5}{2} sin (2 x) - \frac{3}{2} cos (2 x) = 0

化简

\frac{5}{2} sin (2 x) - \frac{3}{2} cos (2 x) : \frac{5 sin ( 2 x ) - 3 cos ( 2 x )}{2}

\frac{5}{2} sin (2 x) - \frac{3}{2} cos (2 x)

乘

\frac{5}{2} sin (2 x) : \frac{5 sin ( 2 x )}{2}

\frac{5}{2} sin (2 x)

分式相乘:

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

= \frac{5 sin ( 2 x )}{2}

= \frac{5 sin ( 2 x )}{2} - \frac{3}{2} cos (2 x)

乘

\frac{3}{2} cos (2 x) : \frac{3 cos ( 2 x )}{2}

\frac{3}{2} cos (2 x)

分式相乘:

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

= \frac{3 cos ( 2 x )}{2}

= \frac{5 sin ( 2 x )}{2} - \frac{3 cos ( 2 x )}{2}

使用法则

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

= \frac{5 sin ( 2 x ) - 3 cos ( 2 x )}{2}

\frac{5 sin ( 2 x ) - 3 cos ( 2 x )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

5 sin (2 x) - 3 cos (2 x) = 0

使用三角恒等式改写

5 sin (2 x) - 3 cos (2 x) = 0

在两边除以

cos (2 x), cos (2 x) \neq = 0

\frac{5 sin ( 2 x ) - 3 cos ( 2 x )}{cos ( 2 x )} = \frac{0}{cos ( 2 x )}

化简

\frac{5 sin ( 2 x )}{cos ( 2 x )} - 3 = 0

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

5 tan (2 x) - 3 = 0

5 tan (2 x) - 3 = 0

将

3

到右边

5 tan (2 x) - 3 = 0

两边加上

3

5 tan (2 x) - 3 + 3 = 0 + 3

化简

5 tan (2 x) = 3

5 tan (2 x) = 3

两边除以

5

5 tan (2 x) = 3

两边除以

5

\frac{5 tan ( 2 x )}{5} = \frac{3}{5}

化简

tan (2 x) = \frac{3}{5}

tan (2 x) = \frac{3}{5}

使用反三角函数性质

tan (2 x) = \frac{3}{5}

tan (2 x) = \frac{3}{5}

的通解

tan (x) = a \Rightarrow x = arctan (a) + 18 0^{\circ} n

2 x = arctan (\frac{3}{5}) + 18 0^{\circ} n

2 x = arctan (\frac{3}{5}) + 18 0^{\circ} n

解

2 x = arctan (\frac{3}{5}) + 18 0^{\circ} n : x = \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

2 x = arctan (\frac{3}{5}) + 18 0^{\circ} n

两边除以

2

2 x = arctan (\frac{3}{5}) + 18 0^{\circ} n

两边除以

2

\frac{2 x}{2} = \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

化简

x = \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

x = \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

x = \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

以小数形式表示解

x = \frac{0.33347 \dots}{2} + \frac{18 0 ^{\circ} n}{2}

作图

流行的例子

tan (θ) = \frac{15}{25}

tan (φ) = 3

sin(x)= 1/(sqrt(2)),0<= x<= 2pi

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

27^2=56^2+35^2-(2*56*35*cos(x))

2 7^{2} = 5 6^{2} + 3 5^{2} - (2 \cdot 56 \cdot 35 \cdot cos (x))

solvefor x,sin^2(x)-sin(x)-2=0

so l v e f or x, sin^{2} (x) - sin (x) - 2 = 0