y=sin(x)+3cos(x) eching | Microsoft math hal qiluvchi

x uchun yechish (complex solution)

x=-i\ln(\frac{\left(-3-i\right)\sqrt{y^{2}-10}+\left(3+i\right)y}{10})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}

x=-i\ln(\left(\frac{3}{10}+\frac{1}{10}i\right)\left(\sqrt{y^{2}-10}+y\right))+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}

y uchun yechish (complex solution)

y=\left(\frac{3}{2}-\frac{1}{2}i\right)e^{ix}+\left(\frac{3}{2}+\frac{1}{2}i\right)e^{-ix}

y uchun yechish

y=\sin(x)+3\cos(x)

Grafik

Viktorina

Trigonometry

5xshash muammolar:

y = \sin ( x ) +3 \cos ( x )

Veb-qidiruvdagi o'xshash muammolar

How do you find the derivative of \displaystyle{y}={2}{\sin{{x}}}+{3}{\cos{{x}}} ?

https://socratic.org/questions/how-do-you-find-the-derivative-of-y-2sinx-3cosx

\displaystyle\text{}\frac{{y}}{{\left.{d}{x}\right.}}={2}{\cos{{x}}}-{3}{\sin{{x}}} Explanation: \displaystyle\text{using the }\ \text{standard derivatives} \displaystyle•{\left({x}\right)}\frac{{d}}{{\left.{d}{x}\right.}}{\left({\sin{{x}}}\right)}={\cos{{x}}}\ \text{ and }\ \frac{{d}}{{\left.{d}{x}\right.}}{\left({\cos{{x}}}\right)}=-{\sin{{x}}} ...

How do you plot the graph of y = \sin(x)+\cos(x) by converting it to another form?

https://math.stackexchange.com/questions/1462034/how-do-you-plot-the-graph-of-y-sinx-cosx-by-converting-it-to-another-f

Special Case It's not that hard. You should just use the summation formula for sines: \sin (x + y) = \sin (x)\cos (y) + \cos (x)\sin (y) This is how it works \eqalign{ \sin (x) + \cos (x) &= \sqrt 2 \left( {{1 \over {\sqrt 2 }}\cos (x) + {1 \over {\sqrt 2 }}\sin (x)} \right) \cr &= \sqrt 2 \left( {\sin ({\pi \over 4})\cos (x) + \cos ({\pi \over 4})\sin (x)} \right) \cr &= \sqrt 2 \sin (x + {\pi \over 4}) \cr} ...

How do you find the derivative \displaystyle{y}={x}{\sin{{x}}}+{\cos{{x}}} ?

https://socratic.org/questions/how-do-you-find-the-derivative-y-xsinx-cosx

Michael Feb 26, 2015 \displaystyle{y}'={x}{\cos{{x}}} Use the product rule for \displaystyle{d}\frac{{{\left({x}{\sin{{x}}}\right)}}}{{{\left.{d}{x}\right.}}} : \displaystyle{\left({u}{v}\right)}'={u}\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}+{v}\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}} ...

How do you simplify \displaystyle{\cos{{\left({x}-{y}\right)}}}+{\sin{{\left({x}-{y}\right)}}} to trigonometric functions of x and y?

https://socratic.org/questions/how-do-you-simplify-cos-x-y-sin-x-y-to-trigonometric-functions-of-x-and-y

= cos y(sin x + cos x) + sin y(sin x - cos x) Explanation: cos (x - y) = cos x.cos y + sin x.sin y sin (x - y) = sin x.cos y - sin y.cos x cos (x - y) + sin (x - y) = cos y(sin x + cos x) + sin y(sin ...

How do you simplify \displaystyle{\sin{{\left({x}+{y}\right)}}}-{\cos{{\left({x}-{y}\right)}}} to trigonometric functions of x and y?

https://socratic.org/questions/how-do-you-simplify-sin-x-y-cos-x-y-to-trigonometric-functions-of-x-and-y

Explanation: It is \displaystyle{\sin{{\left({x}+{y}\right)}}}-{\cos{{\left({x}-{y}\right)}}}={\sin{{x}}}\cdot{\cos{{y}}}+{\cos{{x}}}\cdot{\sin{{y}}}-{\cos{{x}}}\cdot{\cos{{y}}}-{\sin{{x}}}\cdot{\sin{{y}}}={\cos{{x}}}{\left({\sin{{y}}}-{\cos{{y}}}\right)}+{\sin{{x}}}{\left({\cos{{y}}}-{\sin{{y}}}\right)}={\left({\cos{{x}}}-{\sin{{x}}}\right)}{\left({\sin{{y}}}-{\cos{{y}}}\right)}

How do you implicitly differentiate \displaystyle{y}{\sin{{2}}}{x}={x}{\cos{{2}}}{y} ?

https://socratic.org/questions/how-do-you-implicitly-differentiate-y-sin-2x-x-cos-2y

It is \displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=\frac{{{\cos{{2}}}{y}-{2}{y}{\cos{{2}}}{x}}}{{{\sin{{2}}}{x}+{2}{x}{\sin{{2}}}{y}}} Explanation: It is \displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\cdot{\sin{{2}}}{x}+{2}\cdot{y}\cdot{\cos{{2}}}{x}={\cos{{2}}}{y}-{2}{x}{\sin{{2}}}{y}\cdot\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\Rightarrow\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\cdot{\left[{\sin{{2}}}{x}+{2}{x}{\sin{{2}}}{y}\right]}={\cos{{2}}}{y}-{2}{y}{\cos{{2}}}{x}\Rightarrow\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=\frac{{{\cos{{2}}}{y}-{2}{y}{\cos{{2}}}{x}}}{{{\sin{{2}}}{x}+{2}{x}{\sin{{2}}}{y}}}

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