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x=\pi n_{1}+\arctan(2)\text{, }n_{1}\in \mathrm{Z}
x=\pi n_{2}+\pi -\arctan(2)\text{, }n_{2}\in \mathrm{Z}
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5 problemas similares a:
{ \tan ( x ) } ^ {2} = 4
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How do you find the derivative of \displaystyle{\left({1}-{\tan{{x}}}\right)}^{{2}} ?
https://socratic.org/questions/how-do-you-find-the-derivative-of-1-tanx-2
Derivative of \displaystyle{\left({1}-{\tan{{x}}}\right)}^{{2}} is \displaystyle-{2}{{\sec}^{{2}}{x}}+{2}{\tan{{x}}}{{\sec}^{{2}}{x}} Explanation: We can use Chain rule here. Let \displaystyle{f{{\left({x}\right)}}}={\left({1}-{\tan{{x}}}\right)}^{{2}} ...
How do you multiply and simplify \displaystyle{\left({1}+{\tan{{x}}}\right)}^{{2}} ?
https://socratic.org/questions/how-do-you-multiply-and-simplify-1-tanx-2
see below Explanation: \displaystyle{\left({1}+{\tan{{x}}}\right)}^{{2}}={\left({1}+{\tan{{x}}}\right)}{\left({1}+{\tan{{x}}}\right)} ---> FOIL \displaystyle={1}+{\tan{{x}}}+{\tan{{x}}}+{{\tan}^{{2}}{x}} ...
How to integrate (x+\tan x)^2
https://www.quora.com/How-do-I-integrate-x-tan-x-2
Open the brackets. You then have three separate integrals. The first \int x^2dx is simple and equal to \frac {x^3}{3}. The second \int\tan^2xdx is also simple if you remember that \frac {d (\tan x)}{dx}=1+\tan^{2}x ...
Deducing the series expansion of \arctan(x^2) via the series expansion of \arctan(x) at x=0
https://math.stackexchange.com/questions/1652236/deducing-the-series-expansion-of-arctanx2-via-the-series-expansion-of-ar
This approach is perfectly valid. When we have a series \sum_{n=0}^\infty a_nx^n then replacing x\mapsto x^2 we get \sum_{n=0}^\infty a_nx^{2n}=\sum_{n=0}^\infty b_nx^n which is a power ...
\displaystyle{{\tan}^{{2}}{\left({x}\right)}}={0} How can you solve for \displaystyle{x} ?
https://socratic.org/questions/tan-2-x-0-how-can-you-solve-for-x
\displaystyle{x}={k}\pi,{k}\in{Z} Explanation: \displaystyle{{\tan}^{{2}}{x}}={0}\Rightarrow{\left({\tan{{x}}}\right)}^{{2}}={0}\Rightarrow{\tan{{x}}}={0}\Rightarrow{\sin{{x}}}={0} \displaystyle\Rightarrow{x}={k}\pi,{k}\in{Z}
How many solutions does a trigonometric function have 0\le x \le 2\pi?
https://math.stackexchange.com/questions/2118471/how-many-solutions-does-a-trigonometric-function-have-0-le-x-le-2-pi
I do one, you do the other: \tan^22x=1\iff \tan 2x=\pm1\iff 2x=\pm\frac\pi4+k\pi\;,\;\;k\in\Bbb Z\iff \iff x=\pm\frac\pi8+k\frac\pi2\;,\;\;k\in\Bbb Z Hint for the other: \sin3x=-\frac14\iff3x=\arcsin\left(-\frac14\right)+2k\pi\ldots\ldots\text{etc.}
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Problemas similares
\cos ( 3x + \pi ) = 0.5
\sin ( x ) = 1
\sin ( x ) - cos ( x ) = 0
\sin ( x ) + 2 = 3
{ \tan ( x ) } ^ {2} = 4
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