Beint í aðalefni
Microsoft
|
Math Solver
Leysa
Æfing
Leika
Efnisatriði
For-Algebra
Meina
Hamur
Stærsti sameiginlegi þátturinn
Minnst Algengar Margfeldi
Röð aðgerða
Broti
Blandað brot
Prime Factorization
Veldisvísir
Róttækur
Algebra
Sameina samsvarandi hugtök
Leysa fyrir breytu
Þáttur
Rýmka
Metið brot
Línulegar jöfnur
Annars stigs jöfnur
Ójöfnuður
Kerfi jöfnur
Fylki
Hornafræði
Einfalda
Meta
Myndrit
Leysa jöfnur
Stærðfræðigreining
Afleiður
Heildir
Takmörk
Algebra inntak
Inntak hornafræði
Stærðfræðigreining Inntak
Fylkisinntak
Leysa
Æfing
Leika
Efnisatriði
For-Algebra
Meina
Hamur
Stærsti sameiginlegi þátturinn
Minnst Algengar Margfeldi
Röð aðgerða
Broti
Blandað brot
Prime Factorization
Veldisvísir
Róttækur
Algebra
Sameina samsvarandi hugtök
Leysa fyrir breytu
Þáttur
Rýmka
Metið brot
Línulegar jöfnur
Annars stigs jöfnur
Ójöfnuður
Kerfi jöfnur
Fylki
Hornafræði
Einfalda
Meta
Myndrit
Leysa jöfnur
Stærðfræðigreining
Afleiður
Heildir
Takmörk
Algebra inntak
Inntak hornafræði
Stærðfræðigreining Inntak
Fylkisinntak
Frum
algebra
hornafræði
Stærðfræðigreining
tölfræði
Fylki
Stafir
Leystu fyrir x
x=\pi n_{1}+\frac{\pi }{4}
n_{1}\in \mathrm{Z}
Graf
Teikna báðar hliðar í tvívídd
Teikna í tvívídd
Spurningakeppni
Trigonometry
\sin ( x ) = \cos ( x )
Svipuð vandamál úr vefleit
How to solve equations like 2 \sin(x) = \cos(x)
https://math.stackexchange.com/questions/1476944/how-to-solve-equations-like-2-sinx-cosx/1476973
One way can be using tan\frac x2=t so sin x=\frac{2t}{1+t^2} and cos x=\frac{1-t^2}{1+t^2}. Here 2sin x= cos x implies t^2+4t-1=0 from wich tan \frac x2=2\pm\sqrt{5}.Hence the answer of ...
How do you show that the equation \displaystyle{1}-{\sin{{x}}}={\cos{{x}}} is not an identity?
https://socratic.org/questions/how-do-you-show-that-the-equation-1-sinx-cosx-is-not-an-identity
Bdub Nov 12, 2016 Pick a value for x like \displaystyle\frac{\pi}{{3}} and plug it in to both side to show that they don't equal each other and therefore not an identity
How do you solve \displaystyle{1}+{\sin{{\left({x}\right)}}}={\cos{{\left({x}\right)}}} ?
https://socratic.org/questions/how-do-you-solve-1-sin-x-cos-x
\displaystyle{x}={0} Explanation: \displaystyle{1}+{\sin{{\left({x}\right)}}}={\cos{{\left({x}\right)}}}{\quad\text{or}\quad}{\cos{{x}}}-{\sin{{x}}}={1} . Squaring both sides we get \displaystyle{\left({\cos{{x}}}-{\sin{{x}}}\right)}^{{2}}={1}{\quad\text{or}\quad}{{\cos}^{{2}}{x}}+{{\sin}^{{2}}{x}}-{2}{\sin{{x}}}{\cos{{x}}}={1}{\quad\text{or}\quad}{1}-{\sin{{2}}}{x}={1}{\quad\text{or}\quad}{\sin{{2}}}{x}={0}={\sin{{0}}}; ...
Trigonometric equation \sin2x=\cos x
https://math.stackexchange.com/questions/3008492/trigonometric-equation-sin2x-cos-x
As @Nicholas Stull hinted, you lost solutions by not making sure that you were not dividing by zero. As @Winther pointed out, you can avoid this error by factoring. As @Nicholas Stull pointed out, ...
Is there a deeper understanding of the derivative of sin(x) = cos(x)?
https://math.stackexchange.com/q/2454114
Apropos "deeper way": 1) f(x) = f(-x), even fct. Examples: y=x^2, y=cos(x) f'(x) = -f'(-x), chain rule, odd fct. 2) f(x)=-f(-x), odd fct. Examples: y=x^3, y=sin(x). f'(x) = f'(-x), ...
Maximum area of a rectangle inscribed in the cos(x) function
https://math.stackexchange.com/q/2212333
Equations like x= \cos x or x=\cot x generally don't have algebraic solutions. As such, we would first want to note that such an x exists (e.g., by the Intermediate Value Theorem) and then use ...
Meira Vörur
Deila
Afrit
Afritað á klemmuspjald
Svipuð vandamál
\tan ( x )
\sec ( x )
\sin ( x ) = \cos ( x )
\cot ( x )
\cos ( x )
\csc ( x )
Efst á síðu