Why for sin x + cos x = 0 we cannot convert the sinx to cos x tan x and factorise it out like cos x (tan x + 1) = 0. But for 3 tan x = 2 sin x we can?

Question

ganeshie8 · Answer

In the second problem, you get solutions :  
3sin/cos = 2sin
3sin = 2sin cos

sin(3-2cos) = 0

sin = 0,  or cos = 3/2

Notice that cos is NOT 0, when u get the other solution.

ganeshie8 · Answer

For the first problem, you get solutions :
sin x + cos x = 0
cos (tan + 1) = 0.

cos = 0, or tan = -1

Notice that cos = 0 is a part of solution, so you literally dividing ur given equation by 0, which is illegal. so you cannot divide cos in first problem.

anonymous · Answer

But for 3 tan x = 2 sin x if we move the 2 sin x to the left then 3 tan x - 2 sin x is equal to zero.

kc_kennylau · Answer

I wouldn't solve the equation this way.

ganeshie8 · Answer

yes, so ?

ganeshie8 · Answer

you need to be careful when u divide a function -  u must watch out when the stuff you're dividing both sides becomes 0... otherwise u wil get solutions that make no sense

ganeshie8 · Answer

3 tan x - 2 sin x =  0


you can divide any function, as long as it doesn't become 0 in the domain of solutions for this equation.

ganeshie8 · Answer

If you want to divide cos,  may solve like below :

$\large  \sin x + \cos x = 0$

$\large  \sin x \times \dfrac{\cos x}{\cos x} + \cos x = 0$   $\color{red}{\cos x \ne  0}$

$\large  \cos x \times \tan x + \cos x = 0$   $\color{red}{\cos x \ne  0}$

$\large  \cos x (\tan x + 1) = 0$   $\color{red}{\cos x \ne  0}$

$\large  \cos x = 0,   ~  \tan x = -1$   $\color{red}{\cos x \ne  0}$

ganeshie8 · Answer

discard  cosx = 0 solution, pick only tanx = -1 part

anonymous · Answer

Okay ! I got it thanksss!!

ganeshie8 · Answer

np :) good question ! made me think lol.. thanks for asking too :)