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ແກ້
ການປະຕິບັດ
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ປັດໄຈທົ່ວໄປທີ່ຍິ່ງໃຫຍ່ທີ່ສຸດ
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ແກ້ສຳລັບ x
x=\pi n_{1}+\frac{\pi }{4}
n_{1}\in \mathrm{Z}
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Trigonometry
\sin ( x ) = \cos ( x )
ບັນຫາທີ່ຄ້າຍຄືກັນຈາກWeb Search
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 ...
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