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Question Number 208377 by hardmath last updated on 14/Jun/24

sin x − sin (π/6) > 0  x = ?

$$\mathrm{sin}\:\mathrm{x}\:−\:\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\:>\:\mathrm{0} \\ $$$$\mathrm{x}\:=\:? \\ $$

Commented by hardmath last updated on 14/Jun/24

  Solve the inequality

$$ \\ $$Solve the inequality

Answered by lepuissantcedricjunior last updated on 14/Jun/24

x est definie dans quoi!  R ou ]−2𝛑;2𝛑]

$$\boldsymbol{{x}}\:\boldsymbol{{est}}\:\boldsymbol{{definie}}\:\boldsymbol{{dans}}\:\boldsymbol{{quoi}}! \\ $$$$\left.\mathbb{R}\left.\:\boldsymbol{{ou}}\:\right]−\mathrm{2}\boldsymbol{\pi};\mathrm{2}\boldsymbol{\pi}\right] \\ $$

Answered by mr W last updated on 14/Jun/24

sin x > sin (π/6)  ⇒2kπ+(π/6)<x<(2k+1)π−(π/6)

$$\mathrm{sin}\:{x}\:>\:\mathrm{sin}\:\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{6}}<{x}<\left(\mathrm{2}{k}+\mathrm{1}\right)\pi−\frac{\pi}{\mathrm{6}} \\ $$

Commented by hardmath last updated on 14/Jun/24

thankyou dear professor  (π/6) +2πk   and   ((5π)/6) + 2πk ?

$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professor} \\ $$$$\frac{\pi}{\mathrm{6}}\:+\mathrm{2}\pi\mathrm{k}\:\:\:\mathrm{and}\:\:\:\frac{\mathrm{5}\pi}{\mathrm{6}}\:+\:\mathrm{2}\pi\mathrm{k}\:? \\ $$

Commented by mr W last updated on 14/Jun/24

is this not the same as i said?  ((5π)/6)=π−(π/6) !

$${is}\:{this}\:{not}\:{the}\:{same}\:{as}\:{i}\:{said}? \\ $$$$\frac{\mathrm{5}\pi}{\mathrm{6}}=\pi−\frac{\pi}{\mathrm{6}}\:! \\ $$

Commented by hardmath last updated on 14/Jun/24

Dear professor,  k = 0 ⇒ π − (π/6) = ((5π)/6) ?

$$\mathrm{Dear}\:\mathrm{professor}, \\ $$$$\mathrm{k}\:=\:\mathrm{0}\:\Rightarrow\:\pi\:−\:\frac{\pi}{\mathrm{6}}\:=\:\frac{\mathrm{5}\pi}{\mathrm{6}}\:? \\ $$

Commented by mr W last updated on 14/Jun/24

yes

$${yes} \\ $$

Commented by hardmath last updated on 14/Jun/24

thankyou dear professor

$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professor} \\ $$

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