y ପାଇଁ ସମାଧାନ କରନ୍ତୁ
\left\{\begin{matrix}y>0\text{, }&\exists n_{2}\in \mathrm{Z}\text{ : }\beta =2\pi n_{2}+2\pi -\arcsin(\frac{3}{5})\text{ or }\left(\exists n_{4}\in \mathrm{Z}\text{ : }\left(\beta >\frac{\pi \left(4n_{4}+3\right)}{2}\text{ and }\beta <\frac{\pi \left(4n_{4}+5\right)}{2}\right)\text{ and }\exists n_{3}\in \mathrm{Z}\text{ : }\beta =2\pi n_{3}+\pi -\arcsin(\frac{3}{5})\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\beta =\frac{\pi \left(2n_{1}+1\right)}{2}\right)\\y<0\text{, }&\exists n_{5}\in \mathrm{Z}\text{ : }\left(\beta >\frac{\pi \left(4n_{5}+1\right)}{2}\text{ and }\beta <\frac{\pi \left(4n_{5}+3\right)}{2}\right)\text{ and }\exists n_{3}\in \mathrm{Z}\text{ : }\beta =2\pi n_{3}+\pi -\arcsin(\frac{3}{5})\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\beta =\frac{\pi \left(2n_{1}+1\right)}{2}\end{matrix}\right.
β ପାଇଁ ସମାଧାନ କରନ୍ତୁ
\beta \neq \pi n_{1}+\frac{\pi }{2}
\forall n_{1}\in \mathrm{Z}
କ୍ୱିଜ୍
Trigonometry
5 ଟି ପ୍ରଶ୍ନ ଏହି ପରି ଅଟେ:
\tan ( \beta ) = - 3 / 4 , y \cos ( \beta ) > 0 \beta . D
ଅଂଶୀଦାର
କ୍ଲିପ୍ ବୋର୍ଡ଼ରେ ନକଲ କରାଯାଇଛି
ଉଦାହରଣଗୁଡ଼ିକ
ଚତୁଷ୍ପଦୀ ସମୀକରଣ
{ x } ^ { 2 } - 4 x - 5 = 0
ତ୍ରିକୋଣମିତି
4 \sin \theta \cos \theta = 2 \sin \theta
ରୈଖିକ ସମୀକରଣ
y = 3x + 4
ବୀଜଗଣିତ
699 * 533
ମାଟ୍ରିକ୍ସ୍
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
ସମକାଳୀନ ସମୀକରଣ
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
ବିଭେଦୀକରଣ
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
ଇଣ୍ଟିଗ୍ରେସନ୍
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
ସୀମାଗୁଡ଼ିକ
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}