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受欢迎的三角函数 >

28cos(t)=20,0<= t<360

解答

28 cos (t) = 20, 0^{\circ} \leq t < 36 0^{\circ}

解答

t = 0.77519 \dots, t = 36 0^{\circ} - 0.77519 \dots

+1

弧度

t = 0.77519 \dots, t = 2 π - 0.77519 \dots

求解步骤

28 cos (t) = 20, 0^{\circ} \leq t < 36 0^{\circ}

两边除以

28

28 cos (t) = 20

两边除以

28

\frac{28 cos ( t )}{28} = \frac{20}{28}

化简

cos (t) = \frac{5}{7}

cos (t) = \frac{5}{7}

使用反三角函数性质

cos (t) = \frac{5}{7}

cos (t) = \frac{5}{7}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

t = arccos (\frac{5}{7}) + 36 0^{\circ} n, t = 36 0^{\circ} - arccos (\frac{5}{7}) + 36 0^{\circ} n

t = arccos (\frac{5}{7}) + 36 0^{\circ} n, t = 36 0^{\circ} - arccos (\frac{5}{7}) + 36 0^{\circ} n

在

0 \leq t < 36 0^{\circ}

范围内的解

t = arccos (\frac{5}{7}), t = 36 0^{\circ} - arccos (\frac{5}{7})

以小数形式表示解

t = 0.77519 \dots, t = 36 0^{\circ} - 0.77519 \dots

作图

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