2left(sin(x)right)^2left(cos(x)right)^2=1

x üçün həll et (complex solution)

x=2\pi n_{9}+\frac{5}{4}\pi +\frac{1}{2}i\ln(2)+\left(-\frac{1}{2}i\right)\ln(-2+2^{\frac{3}{2}})\text{, }n_{9}\in \mathrm{Z}

x=2\pi n_{9}+\frac{7}{4}\pi +\frac{1}{2}i\ln(2)+\left(-\frac{1}{2}i\right)\ln(-2+2^{\frac{3}{2}})\text{, }n_{9}\in \mathrm{Z}

x=\frac{1}{4}\pi +\frac{1}{2}i\ln(2)+2\pi n_{28}+\left(-\frac{1}{2}i\right)\ln(-2+2^{\frac{3}{2}})\text{, }n_{28}\in \mathrm{Z}

x=\frac{3}{4}\pi +\frac{1}{2}i\ln(2)+2\pi n_{28}+\left(-\frac{1}{2}i\right)\ln(-2+2^{\frac{3}{2}})\text{, }n_{28}\in \mathrm{Z}

x=2\pi n_{41}+\frac{5}{4}\pi +\frac{1}{2}i\ln(2)+\left(-\frac{1}{2}i\right)\ln(2+2^{\frac{3}{2}})\text{, }n_{41}\in \mathrm{Z}

x=2\pi n_{41}+\frac{7}{4}\pi +\frac{1}{2}i\ln(2)+\left(-\frac{1}{2}i\right)\ln(2+2^{\frac{3}{2}})\text{, }n_{41}\in \mathrm{Z}

x=\frac{1}{2}i\ln(2)+\frac{1}{4}\pi +\left(-\frac{1}{2}i\right)\ln(2+2^{\frac{3}{2}})+2\pi n_{48}\text{, }n_{48}\in \mathrm{Z}

x=\frac{1}{2}i\ln(2)+\frac{3}{4}\pi +\left(-\frac{1}{2}i\right)\ln(2+2^{\frac{3}{2}})+2\pi n_{48}\text{, }n_{48}\in \mathrm{Z}

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Sorğu

Trigonometry

5 oxşar problemlər:

2 { \left( \sin ( x ) \right) }^{ 2 } { \left( \cos ( x ) \right) }^{ 2 } =1

Veb Axtarışdan Oxşar Problemlər

Can we prove \sin(x)^2 + \cos(x)^2 = 1 using just their series?

https://math.stackexchange.com/q/2689947

Absolutely. Power series actually multiply just like polynomials do: (a_0+a_1x+a_2x^2+a_3x^3+\ldots)(b_0+b_1x+b_2x^2+\ldots)=\sum_{n=0}^{\infty}\left(\sum_{c=0}^na_cb_{n-c}\right)x^n. Let \alpha(x)=\left(\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!}\right)^2 ...

How do you differentiate \displaystyle{f{{\left({x}\right)}}}={\sin{{\left({x}^{{2}}\right)}}}{\cos{{\left({x}^{{2}}\right)}}} using the product rule?

https://socratic.org/questions/how-do-you-differentiate-f-x-sin-x-2-cos-x-2-using-the-product-rule

\displaystyle{f}'{\left({x}\right)}={2}{x}{\cos{{\left({2}{x}^{{2}}\right)}}} Explanation: \displaystyle{f}'{\left({x}\right)}={\cos{{\left({x}^{{2}}\right)}}}\frac{{d}}{{\left.{d}{x}\right.}}{\left[{\sin{{\left({x}^{{2}}\right)}}}\right]}+{\sin{{\left({x}^{{2}}\right)}}}\frac{{d}}{{\left.{d}{x}\right.}}{\left[{\cos{{\left({x}^{{2}}\right)}}}\right]} ...

What is the maximum value of 4(\sin x)^2 + 3(\cos x)^2

https://math.stackexchange.com/questions/552838/what-is-the-maximum-value-of-4-sin-x2-3-cos-x2

Your argument is OK. You can check it taking the derivative of the function: f(x)=4\sin(x)^2+3\cos(x)^2 y'(x)=2\sin(x)\cos(x) which is null for x=0,x=\frac{\pi}{2} The first value gives y(x)=3 ...

How do you differentiate \displaystyle{\sin{{\left({x}^{{2}}\right)}}}{\left({\cos{{\left({x}^{{2}}\right)}}}\right)} ?

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

\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}{\sin{{\left({x}^{{2}}\right)}}}{\cos{{\left({x}^{{2}}\right)}}}={2}{x}{\cos{{\left({2}{x}^{{2}}\right)}}} Explanation: This problem can also be ...

How do you solve \displaystyle{2}{{\cos}^{{2}}{x}}+{3}{\sin{{x}}}={3} in the interval \displaystyle{0}\le{x}\le{2}\pi ?

https://socratic.org/questions/how-do-you-solve-2cos-2x-3sinx-3-in-the-interval-0-x-2pi

Solution \displaystyle{x}=\frac{\pi}{{6}},\frac{\pi}{{2}},\frac{{{5}\pi}}{{6}} for \displaystyle{0}\le{x}\le{2}\pi Explanation: \displaystyle{2}{{\cos}^{{2}}{x}}+{3}{\sin{{x}}}={3}{\quad\text{or}\quad}{2}{\left({1}-{{\sin}^{{2}}{x}}\right)}+{3}{\sin{{x}}}-{3}={0}{\quad\text{or}\quad}{2}{{\sin}^{{2}}{x}}-{3}{\sin{{x}}}+{1}={0}{\quad\text{or}\quad}{\left({2}{\sin{{x}}}-{1}\right)}{\left({\sin{{x}}}-{1}\right)}={0} ...

How do you solve? \displaystyle{2}{\left({\sin{{x}}}\right)}^{{2}}−{\cos{{x}}}−{1}={0}

https://socratic.org/questions/how-do-you-solve-2-sin-x-2-cos-x-1-0

Douglas K. May 15, 2018 Given: \displaystyle{2}{{\sin}^{{2}}{\left({x}\right)}}−{\cos{{\left({x}\right)}}}−{1}={0} Substitute: \displaystyle{{\sin}^{{2}}{\left({x}\right)}}={1}-{{\cos}^{{2}}{\left({x}\right)}} ...

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