1. ## Solving equations

Hi
Need help solving $cot(x)+3tan(x)=5cosec(x)$, i have tried to do the following:

$\frac{1}{tan(x)+3tan(x)}=\frac{5}{sin(x)}$

$\frac{cos(x)}{sin(x)}+\frac{3sin(x)}{cos(x)}=\frac {5}{sin(x)}$

$\frac{cos^2(x)+3sin^2(x)}{sin(x)cos(x)}=\frac{5cos (x)}{sin(x)cos(x)}$

after this i contiue and i couldn't find the solution.

P.S

2. $\cot(x)+3\tan(x)\neq \frac{1}{\tan(x)+3\tan(x)}$

Try

$
\cot(x)+3\tan(x)= \frac{1}{\tan(x)}+3\tan(x)$

3. $
\frac{cos^2(x)+3sin^2(x)=5cos(x)}{sin(x)cos(x)}
$

$1-sin^2(x)+3sin^2(x)=5cos(x)$

$2sin^2(x)=5cos(x)-1$

$1-cos^2(x)=5cos(x)-1$

$2=5cos(x)+cos^2(x)$

$2=cos(x)(5+cos(x))$

so cos(x)=2 and 5+cos(x)=2

Someone tell me where i have gone wrong.

4. There is something wrong with your first equation. You cannot divide an equation. You need to put the fraction on both sides :

$\frac{cos^2(x)+3sin^2(x)}{sin(x)cos(x)}=\frac{5cos (x)}{sin(x)cos(x)}$

This simplifies trivially to :

$cos^2(x)+3sin^2(x) = 5cos(x)$

Now you can use the trigonometric identity : $cos^2(x) + sin^2(x) = 1$

$(1 - sin^2(x) )+3sin^2(x) = 5cos(x)$

$1 + 2sin^2(x) = 5cos(x)$

$1 + 2sin^2(x) = 5cos(x)$

Use this again : $cos^2(x) + sin^2(x) = 1$

$1 + 2(1 - cos^2(x)) = 5cos(x)$

$1 + 2 - 2 cos^2(x)) = 5cos(x)$

$3 - 2 cos^2(x)) = 5cos(x)$

$2 cos^2(x) + 5cos(x) - 3 = 0$

And you are left with a quadratic equation, which can be easily solved for $cos(x)$, and thus for $x$.
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You got it wrong between the third and the fourth line : $2 sin^2 (x) = 2 - 2 cos^2 (x) \neq 1 - cos^2(x)$