# Nghiệm pt sin2x

2cos2x + 2sin2xExplanation:differentiate using \displaystyle\ \text{ chain rule }\ \displaystyle\frac{{d}}{{\left.{d}{x}\right.}}{\left<{f{{\left({g{{\left({x}\right)}}}\right)}}}\right>}={f}'{\left({g{{\left({x}\right)}}}\right)}.{g}'{\left({x}\right)} ...

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https://math.stackexchange.com/questions/1970341/i-need-help-writing-the-expression-5-sin-2x-cos-2x-in-terms-of-sine-only
Given a\sin(u)+b\cos(v), factor out whatever you need to to make the coefficients of \sin and \cos have sum of squares equal to 1. Then a sum of angles formula applies. In this case, 5\sqrt{2}\left(\sin(2x)\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\cos(2x)\right) ...
How do you find the integral of \displaystyle{\left({\sin{{\left({5}{x}\right)}}}\right)}{\left({5}{\cos{{\left({5}{x}\right)}}}\right)}{2} ?
Massimiliano Jan 31, 2015 The answer is:\displaystyle-\frac{{1}}{{2}}{\cos{{10}}}{x}+{c} .The double-angle formula says: \displaystyle{\sin{{\left({2}\alpha\right)}}}={2}{\sin{\alpha}}{\cos{\alpha}}{)} ...

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\displaystyle{x}=\frac{{{3}\pi}}{{8}}+{k}\pi\displaystyle{x}=\frac{{{7}\pi}}{{8}}+{k}\pi Explanation:sin 2x + cos 2x = 0Use trig identity:\displaystyle{\sin{{a}}}+{\cos{{a}}}=\sqrt{{2}}{\cos{{\left({a}-\frac{\pi}{{4}}\right)}}} ...
Hint: Our function is \sqrt{65}\left(\frac{1}{\sqrt{65}}\sin 2x -\frac{8}{\sqrt{65}}\cos 2x\right). Let b be an angle whose cosine is \frac{1}{\sqrt{65}} and whose sine is \frac{8}{\sqrt{65}} ...

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Nghi N. Jun 21, 2018\displaystyle{2}{\sin{{2}}}{x}.{\cos{{2}}}{x}={\sin{{4}}}{x} .To solve for x, it needs to have an equation.For example, solve: \displaystyle{2}{\sin{{2}}}{x}.{\cos{{2}}}{x}={0} ...
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\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>

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