Leysa cos^2xsin^2x-10sinxcos^3x+21cos^4x=0 | Microsoft stærðfræði leysa

Leystu fyrir x (complex solution)

x=\pi n_{1}+\arctan(3)\text{, }n_{1}\in \mathrm{Z}

x=\pi n_{2}+\arctan(7)\text{, }n_{2}\in \mathrm{Z}

x=\pi n_{3}+\frac{\pi }{2}\text{, }n_{3}\in \mathrm{Z}

Leystu fyrir x

x=\pi n_{2}+\arcsin(\frac{7\sqrt{2}}{10})\text{, }n_{2}\in \mathrm{Z}

x=\pi n_{3}+\frac{\pi }{2}\text{, }n_{3}\in \mathrm{Z}

x=\pi n_{1}+\arccos(\frac{\sqrt{10}}{10})\text{, }n_{1}\in \mathrm{Z}

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Spurningakeppni

Trigonometry

\cos ^ { 2 } x \sin ^ { 2 } x - 10 \sin x \cos ^ { 3 } x + 21 \cos ^ { 4 } x = 0

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Trigonometric equation: 2(\sin^6 x+\cos^6 x)-3(\sin^4 x+\cos^4 x)+1=0

https://math.stackexchange.com/questions/598045/trigonometric-equation-2-sin6-x-cos6-x-3-sin4-x-cos4-x1-0

As \sin^2x+\cos^2x=1 Taking square, \displaystyle1=(\sin^2x+\cos^2x)^2=\sin^4x+\cos^4x+2\sin^2x\cos^2x \displaystyle\implies \sin^4x+\cos^4x=1-2(\sin^2x\cos^2x)\ \ \ \ (1) Taking cube, \displaystyle1=(\sin^2x+\cos^2x)^3=\sin^6x+\cos^6x+3\sin^2x\cos^2x(\sin^2x+\cos^2x) ...

How do you verify \displaystyle{{\sin}^{{5}}{x}}{{\cos}^{{2}}{x}}={\left({{\cos}^{{2}}{x}}-{2}{{\cos}^{{4}}{x}}+{{\cos}^{{6}}{x}}\right)}{\sin{{x}}} ?

https://socratic.org/questions/how-do-you-verify-sin-5xcos-2x-cos-2x-2cos-4x-cos-6x-sinx

See below Explanation: Operating in both sides \displaystyle{{\sin}^{{5}}{x}}{{\cos}^{{2}}{x}}={\left({{\cos}^{{2}}{x}}-{2}{{\cos}^{{4}}{x}}+{{\cos}^{{6}}{x}}\right)}{\sin{{x}}} \displaystyle{\sin{{x}}}{{\sin}^{{4}}{x}}{{\cos}^{{2}}{x}}={\left({{\cos}^{{2}}{x}}-{2}{{\cos}^{{4}}{x}}+{{\cos}^{{6}}{x}}\right)}{\sin{{x}}} ...

How do you verify \displaystyle{\left({1}-{{\cos}^{{2}}{x}}\right)}{\left({1}+{{\cos}^{{2}}{x}}\right)}={2}{{\sin}^{{2}}{x}}-{{\sin}^{{4}}{x}} ?

https://socratic.org/questions/how-do-you-verify-1-cos-2x-1-cos-2x-2sin-2x-sin-4x

Prove ( 1 - cos ^2 x) (1 + cos^2 x) = 2sin^2 x - sin^4 x Explanation: Replace \displaystyle{\left({1}-{{\cos}^{{2}}{x}}\right)} by \displaystyle{{\sin}^{{2}}{x}} and \displaystyle{{\cos}^{{2}}{x}} ...

How do you verify \displaystyle{2}+{{\cos}^{{2}}{x}}-{3}{{\cos}^{{4}}{x}}={{\sin}^{{2}}{x}}{\left({2}+{3}{{\cos}^{{2}}{x}}\right)} ?

https://socratic.org/questions/how-do-you-verify-2-cos-2-x-3-cos-4-x-sin-2-x-2-3-cos-2-x

Noah G Mar 8, 2018 Use \displaystyle{{\sin}^{{2}}{x}}={1}-{{\cos}^{{2}}{x}} . RHS: \displaystyle{\left({1}-{{\cos}^{{2}}{x}}\right)}{\left({2}+{3}{{\cos}^{{2}}{x}}\right)} \displaystyle{2}-{2}{{\cos}^{{2}}{x}}+{3}{{\cos}^{{2}}{x}}-{3}{{\cos}^{{4}}{x}} ...

Finding minima and maxima of \sin^2x \cos^2x

https://math.stackexchange.com/questions/477636/finding-minima-and-maxima-of-sin2x-cos2x

It s easier to deal with this form f(x) = \sin^2(x)\cos^2(x)=\frac{\sin^2(2x)}{4} \implies f'(x) = \sin(2x)\cos(2x)=\frac{\sin(4x)}{2}. Now, you should be able to finish the problem.

Integral of Trigonometric Identities

https://math.stackexchange.com/questions/1315757/integral-of-trigonometric-identities

Note f(x)=\sin^3 x \cos^2 2x. You have \cos 2x = 2 \cos^2 x -1 Hence (\sin^3 x)(\cos^2 2x)=\sin x \sin^2 x (2 \cos^2 x -1)=\sin x (1- \cos^2 x) (2 \cos^2 x -1) As (\cos x)^\prime = -\sin x ...

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