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受欢迎的三角函数 >

sin(x)=-(sqrt(3))/5 \land cos(x)>0

解答

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

解答

x = - 0.35374 \dots + 2 πn

+1

间隔符号

x = - 0.35374 \dots + 2 πn

求解步骤

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

sin (x) = - \frac{3}{5} : x = - 0.35374 \dots + 2 πn, x = π + 0.35374 \dots + 2 πn

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

使用反三角函数性质

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

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

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{3}{5}) + 2 πn, x = π + arcsin (\frac{3}{5}) + 2 πn

x = arcsin (- \frac{3}{5}) + 2 πn, x = π + arcsin (\frac{3}{5}) + 2 πn

以小数形式表示解

x = - 0.35374 \dots + 2 πn, x = π + 0.35374 \dots + 2 πn

cos (x) > 0 : - \frac{π}{2} + 2 πn < x < \frac{π}{2} + 2 πn

cos (x) > 0

对于

cos (x) > a

，若

- 1 \leq a < 1

，则

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (0) + 2 πn < x < arccos (0) + 2 πn

化简

- arccos (0) : - \frac{π}{2}

- arccos (0)

使用以下普通恒等式:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{π}{2}

化简

arccos (0) : \frac{π}{2}

arccos (0)

使用以下普通恒等式:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2}

- \frac{π}{2} + 2 πn < x < \frac{π}{2} + 2 πn

合并区间

(x = π + 0.35374 \dots + 2 πn or x = - 0.35374 \dots + 2 πn) and - \frac{π}{2} + 2 πn < x < \frac{π}{2} + 2 πn

合并重叠的区间

x = - 0.35374 \dots + 2 πn

流行的例子

sin(x)0<= x<= pi

sin (x) 0 \leq x \leq π

x=-4\land csc(x)>0

x = - 4 and csc (x) > 0

cos(θ)<0\land (cos(θ))(sin(θ))<0

cos (θ) < 0 and (cos (θ)) (sin (θ)) < 0

cosh(θ)= 15/4 \land θ<0,sinh(θ)

cosh (θ) = \frac{15}{4} and θ < 0, sinh (θ)

2pi>sqrt(3)tan(θ)+1>= 0

2 π > 3 tan (θ) + 1 \geq 0