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7:13 AM
@tilper Do you think that we need radical-equations tag. There already is the tag (radicals). — Martin Sleziak 17 secs ago
-8
Q: Solve this equation: $(7x-10)\sqrt{x-2}=2(1+\sqrt{2x-1})(2\sqrt{2x-3}+\sqrt{x-2})$

Road HumanSolve this equation: $$(7x-10)\sqrt{x-2}=2(1+\sqrt{2x-1})(2\sqrt{2x-3}+\sqrt{x-2})$$ I've created it and i have a nice solution, do you want to solve?

 
 
1 hour later…
8:20 AM
Questions currently tagged :
4
Q: solve $\tan{x} = \tan{3x}$

user360165I'm asked to solve $\tan{x} = \tan{3x}$ Here's my attempt: $$\tan{x} = \tan{3x}$$ $$\tan{x} = \tan{(x + 2x)}$$ $$\tan{x} = \frac{\tan{x} + \tan{2x}}{1-\tan{x}\tan{2x}}$$ Recall the identity: $$\tan{2x} = \frac{2\tan{x}}{1-\tan^2{x}}$$ So then we have: $$\tan{x} = \frac{\tan{x} + \frac{2\tan{...

0
Q: How to solve $a_1 \sin (x+y) + a_2 \sin (x-y)=0$

unseen_riderI'm looking for methods on how to solve; $$a_1 \sin (x+y) + a_2 \sin (x-y)=0$$ where 1) $a_1, a_2$ are constants 2) $a_1, a_2$ are functions of $x$ and $y$

0
Q: Excluding erroneous solutions of a trigonometric equation

Akash BajajConsider the following trigonometric equation, which needs to be solved for $\theta$: $\tan{(\pi\cot{\theta})}=\cot{(\pi\tan{\theta})}$ The solution given is: $\tan{\theta}=\dfrac{2n+1\pm\sqrt{4n^2+4n-15}}{4}$ for integral values of $n$, AND $n>1$ or $n<-2$. The method I adopted to arrive at t...

1
Q: Given $\tan a = \frac{1}{7}$ and $\sin b = \frac{1}{\sqrt{10}}$, show $a+2b = \frac{\pi}{4}$.

MM PP Given $$\tan a = \frac{1}{7} \qquad\text{and}\qquad \sin b = \frac{1}{\sqrt{10}}$$ $$a,b \in (0,\frac{\pi}{2})$$ Show that $$a+2b=\frac{\pi}{4}$$ Does exist any faster method of proving that, other than expanding $\sin{(a+2b)}$? Thank You!

2
A: Tag management 2016

Martin SleziakThe tag trigonometric-equations has been created recently. It is still listed among new tags. At the moment there are 4 questions in that tag. As far as I can say, questions about trigonometric equations have been tagged trigonometry so far. And also trigonometric-functions and trigonometric-ide...

 

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