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受欢迎的三角函数 >

2sin(x)<= 1,2cos(x)<sqrt(3)

解答

2 sin (x) \leq 1, 2 cos (x) < 3

解答

- \frac{7 π}{6} + 2 πn \leq x \leq \frac{π}{6} + 2 πn

+2

间隔符号

[- \frac{7 π}{6} + 2 πn, \frac{π}{6} + 2 πn]

十进制

- 3.66519 \dots + 2 πn \leq x \leq 0.52359 \dots + 2 πn

求解步骤

2 sin (x) \leq 1

两边除以

2

2 sin (x) \leq 1

两边除以

2

\frac{2 sin ( x )}{2} \leq \frac{1}{2}

化简

sin (x) \leq \frac{1}{2}

sin (x) \leq \frac{1}{2}

对于

sin (x) \leq a

，若

- 1 < a < 1

，则

- π - arcsin (a) + 2 πn \leq x \leq arcsin (a) + 2 πn

- π - arcsin (\frac{1}{2}) + 2 πn \leq x \leq arcsin (\frac{1}{2}) + 2 πn

化简

- π - arcsin (\frac{1}{2}) : - \frac{7 π}{6}

- π - arcsin (\frac{1}{2})

使用以下普通恒等式:

arcsin (\frac{1}{2}) = \frac{π}{6}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= - π - \frac{π}{6}

化简

- π - \frac{π}{6}

将项转换为分式:

π = \frac{π 6}{6}

= - \frac{π 6}{6} - \frac{π}{6}

因为分母相等，所以合并分式:

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

= \frac{- π 6 - π}{6}

同类项相加:

- 6 π - π = - 7 π

= \frac{- 7 π}{6}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{7 π}{6}

= - \frac{7 π}{6}

化简

arcsin (\frac{1}{2}) : \frac{π}{6}

arcsin (\frac{1}{2})

使用以下普通恒等式:

arcsin (\frac{1}{2}) = \frac{π}{6}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

- \frac{7 π}{6} + 2 πn \leq x \leq \frac{π}{6} + 2 πn

流行的例子

solvefor θ,3r^2cos^2(θ)+r^2sin^2(θ)<= 1

so l v e f or θ, 3 r^{2} cos^{2} (θ) + r^{2} sin^{2} (θ) \leq 1

(sin(x)-1/2)(sin(x)-7/2)<= 0

(sin (x) - \frac{1}{2}) (sin (x) - \frac{7}{2}) \leq 0

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

tan(1/x)<= tan(1/(x+1))

tan (\frac{1}{x}) \leq tan (\frac{1}{x + 1})

cos(x^4)+sin(x^4)>= 0.5

cos (x^{4}) + sin (x^{4}) \geq 0.5