solve: cosx+1=sinx, [0,2π)

Question

callisto · Answer

$$\cos x+1=\sin x$$$$\cos x - \sin x = -1$$By squaring both sides, what do you get?

zzr0ck3r · Answer

clever:)

callisto · Answer

I'm too stupid to think of other tricks :(

zzr0ck3r · Answer

lol:)

jhannybean · Answer

There is nothing stupid abut that. It's the easiest method.

hartnn · Answer

yup, nice way....
other longer method would be to square both sides initially to get a quadratic in cos^2 x....

zzr0ck3r · Answer

or just look at it:)

anonymous · Answer

... I think I'm really missing it right now because I don't get what seems so easy to you T_T... Haha sorry but if you squared it like so: (cosx-sinx)^2=1, wouldn't it become more complicated?  cosx^2-2cosxsinx+sinx^2=1  What do I do then?

jhannybean · Answer

$$(\cos x - \sin x)^2 = (-1)^2$$

hartnn · Answer

use the most standard identity of trigonometry
$\large \sin^2x+\cos^2x=1$

callisto · Answer

Something you need:
$$cos^2x+sin^2x =1$$$$2sinxcosx=sin(2x)$$

jhannybean · Answer

$$(\cos x - \sin x)^2 = (-1)^2$$$$(\cos x-\sin x)(\cos x - \sin x)=(-1)(-1)$$can you FOIL and simplify?

anonymous · Answer

yeah I did, I understand now. Thanks :)

anonymous · Answer

wait, one last question though. Is there a difference between $$\cos ^{2}x$$ and $$cosx ^{2}$$

callisto · Answer

Yes.

callisto · Answer

$$cos^2x = (cosx)^2$$You take the cosine before taking the square.
$$cosx^2=cos(x^2)$$You take the square before taking the cosine.

anonymous · Answer

Thank you! :) I was planning to use that identity too but I thought that if you squared both side it would become cos(x^2) and that didn't match the identity. :O But thanks for clearing it up!!

callisto · Answer

Welcome~~~~~ :)