sin ⁡ x = a \sin x=a sin x = a тригонометриялық теңдеуінің шешімінің жалпы түрі

x = ( − 1 ) n arcsin ⁡ a + π n , n ∈ Z x=\left(-1\right)^n\arcsin a+\pi n,\ n\in Z x = ( − 1 ) n arcsin a + π n , n ∈ Z

x = ± arccos ⁡ a + 2 π n , n ∈ Z x=\pm\arccos a+2\pi n,\ n\in Z x = ± arccos a + 2 π n , n ∈ Z

x = π 2 + 2 π n , n ∈ Z x=\frac{\pi}{2}+2\pi n,\ n\in Z x = 2 π + 2 π n , n ∈ Z

x = − π 4 + π n , n ∈ Z x=-\frac{\pi}{4}+\pi n,\ n\in Z x = − 4 π + π n , n ∈ Z

cos ⁡ x = a \cos x=a\ cos x = a тригонометриялық теңдеуінің шешімінің жалпы түрі

x = ± arccos ⁡ a + 2 π n , n ∈ Z x=\pm\arccos a+2\pi n,\ n\in Z x = ± arccos a + 2 π n , n ∈ Z

x = arccos ⁡ a + π n , n ∈ Z x=\arccos a+\pi n,\ n\in Z x = arccos a + π n , n ∈ Z

x = arccos ⁡ a + 2 π n , n ∈ Z x=\arccos a+2\pi n,\ n\in Z x = arccos a + 2 π n , n ∈ Z

x = π 2 + 2 π n , n ∈ Z x=\frac{\pi}{2}+2\pi n,\ n\in Z x = 2 π + 2 π n , n ∈ Z

t g x = a tgx=a\ t g x = a теңдеуінің шешімі

x = a r ctg ⁡ a + π n , n ∈ Z x=ar\operatorname{ctg}a+\pi n,\ n\in Z x = a r c t g a + π n , n ∈ Z

x = arcctg ⁡ a + π n , n ∈ Z x=\operatorname{arcctg}a\ +\pi n,\ n\in Z x = a r c c t g a + π n , n ∈ Z

x = a r ctg ⁡ a + 2 π n , n ∈ Z x=ar\operatorname{ctg}a+2\pi n,\ n\in Z x = a r c t g a + 2 π n , n ∈ Z

x = arcctg ⁡ a + 2 π n , n ∈ Z x=\operatorname{arcctg}a\ +2\pi n,\ n\in Z x = a r c c t g a + 2 π n , n ∈ Z

ctg ⁡ x = a \operatorname{ctg}x=a c t g x = a теңдеуінің шешімі

x = arcctg ⁡ a + π n , n ∈ Z x=\operatorname{arcctg}a+\pi n,\ n\in Z x = a r c c t g a + π n , n ∈ Z

x = a r ctg ⁡ a + π n , n ∈ Z x=ar\operatorname{ctg}a+\pi n,\ n\in Z x = a r c t g a + π n , n ∈ Z

x = a r ctg ⁡ a + 2 π n , n ∈ Z x=ar\operatorname{ctg}a+2\pi n,\ n\in Z x = a r c t g a + 2 π n , n ∈ Z

x = arcctg ⁡ a + 2 π n , n ∈ Z x=\operatorname{arcctg}a+2\pi n,\ n\in Z x = a r c c t g a + 2 π n , n ∈ Z

arcctg ⁡ ( − a ) \operatorname{arcctg}\left(-a\right) a r c c t g ( − a )

arcctg ⁡ ( π − a ) \operatorname{arcctg}\left(\pi-a\right) a r c c t g ( π − a )

a r ctg ⁡ ( π − a ) ar\operatorname{ctg}\left(\pi-a\right) a r c t g ( π − a )

− arcctg ⁡ ( − a ) -\operatorname{arcctg}\left(-a\right) − a r c c t g ( − a )

10 сынып АЛГЕБРА 2 тоқсан ТЖБ2

Authored by Maksat Bakytbek

Mathematics

10th Grade

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10 сынып АЛГЕБРА 2 тоқсан ТЖБ2

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