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受欢迎的三角函数 >

sqrt(2)cos(θ)=1,0<= θ<= 360

解答

2 cos (θ) = 1, 0^{\circ} \leq θ \leq 36 0^{\circ}

解答

θ = 4 5^{\circ}, θ = 31 5^{\circ}

+1

弧度

θ = \frac{π}{4}, θ = \frac{7 π}{4}

求解步骤

2 cos (θ) = 1, 0^{\circ} \leq θ \leq 36 0^{\circ}

两边除以

2

2 cos (θ) = 1

两边除以

2

\frac{2 cos ( θ )}{2} = \frac{1}{2}

化简

\frac{2 cos ( θ )}{2} = \frac{1}{2}

化简

\frac{2 cos ( θ )}{2} : cos (θ)

\frac{2 cos ( θ )}{2}

约分:

2

= cos (θ)

化简

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

cos (θ) = \frac{2}{2}

cos (θ) = \frac{2}{2}

cos (θ) = \frac{2}{2}

cos (θ) = \frac{2}{2}

的通解

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}

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

在

0 \leq θ \leq 36 0^{\circ}

范围内的解

θ = 4 5^{\circ}, θ = 31 5^{\circ}

作图

流行的例子

cos (B) = \frac{3}{8}

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

4 sec^{2} (x) = 8

sqrt(3)*sin(x)-cos(x)=sqrt(2)

3 \cdot sin (x) - cos (x) = 2

cos(3x)=sin(60)

cos (3 x) = sin (6 0^{\circ})