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ປັດໄຈທົ່ວໄປທີ່ຍິ່ງໃຫຍ່ທີ່ສຸດ
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ແກ້ສຳລັບ x
x=2\pi n_{1}+\frac{\pi }{2}
n_{1}\in \mathrm{Z}
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5 ບັນຫາທີ່ຄ້າຍຄືກັນກັບ:
\sin ( x ) = 1
ບັນຫາທີ່ຄ້າຍຄືກັນຈາກWeb Search
Particular integral for x\sin(1-x)?
https://math.stackexchange.com/questions/1265354/particular-integral-for-x-sin1-x
Try splitting up \sin(1-x) using the difference formula for \sin - \sin(\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \! This should tell you what your particular ...
Limits of the solutions to x\sin x = 1
https://math.stackexchange.com/questions/719244/limits-of-the-solutions-to-x-sin-x-1
Because you are always evaluating the limit, this is an asymptotic expansion of the explicit expression for the solutions. Write x=2\pi n +\epsilon You get \sin \epsilon=\frac{1}{2\pi n +\epsilon} ...
How to determine the solution to \sin x = 8 in the complex numbers
https://www.quora.com/How-do-you-determine-the-solution-to-sin-x-8-in-the-complex-numbers
Let’s just solve the general \sin x = t and later set t=8. e^{ix}= \cos x + i \sin x e^{-ix} = \cos x - i \sin x e^{ix} - e^{-ix} = 2i \sin x \sin x = \dfrac{e^{ix} - e^{-ix}}{2i} ...
Find the critical points of: y = x + \cos(x)
https://math.stackexchange.com/q/124497
You're essentially there: y = x + \cos(x) = \frac\pi2 + 2\pi k + \cos(\frac\pi2 + 2\pi k) = \frac\pi2 + 2\pi k. There are infinitely many y-values, one for each k \in \mathbb{Z}.
Solve \cos 2x - \sin x = 0 for 0 \le x \le 360
https://math.stackexchange.com/questions/1778416/solve-cos-2x-sin-x-0-for-0-le-x-le-360
From 2 \sin x=1, you should have \sin x=0.5. Sine is positive in the first two quadrants, you should obtain 30^{\circ} and 150^{\circ} as your solution as well.
Trigonometric function solutions within an interval
https://math.stackexchange.com/q/1005511
YOU are right. I cannot find any errors.
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\cos ( 3x + \pi ) = 0.5
\sin ( x ) = 1
\sin ( x ) - cos ( x ) = 0
\sin ( x ) + 2 = 3
{ \tan ( x ) } ^ {2} = 4
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