Решить неравенства:
1) [tex]tgx\ \textgreater \ \frac{1}{ \sqrt{3} } [/tex]
2)[tex]tg2x \geq \frac{1}{ \sqrt{3} } [/tex]
3) [tex]tg \frac{x}{6} \ \textless \ \frac{1}{ \sqrt{3} } [/tex]
4) [tex]tg(x- \frac{ \pi }{9} ) \leq - \frac{1}{ \sqrt{3} } [/tex]
1) [tex]tgx\ \textgreater \ \frac{1}{ \sqrt{3} } [/tex]
2)[tex]tg2x \geq \frac{1}{ \sqrt{3} } [/tex]
3) [tex]tg \frac{x}{6} \ \textless \ \frac{1}{ \sqrt{3} } [/tex]
4) [tex]tg(x- \frac{ \pi }{9} ) \leq - \frac{1}{ \sqrt{3} } [/tex]
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Answers & Comments
tgx>1/√3
x∈(π/6+πk;π/2+πk,k∈z)
2
tg2x≥1/√3
π/6+πk≤2x≤π/2+πk,k∈z
x∈[π/12+πk/2;π/4+∈z)
3
tgx/3<1/√3
-π/2+πk<x/3<π/6+πk,k∈z
-3π/2+3πk<x<π/2+3πk,k∈z
4
tg(x-π/9)≤-1/√3
-π/2+πk≤x-π/9≤-π/6+πk,k∈z
-π/2+π/9+πk≤x≤-π/6+π/9+πk,k∈z
-7π/18+πk≤x≤-π/18+πk,k∈z