Question

find general solution of 3tanx + cotx= 5cosecx

Answer 1

Equating equations we get cos x=1/2 or -3 but cos x ranges from +1 to -1 ,
So, denying x=-3,
We get x=1/2
Cos x =1/2=cos 60°
x=60°

Answer 2

Answer:

The general solution of the expression is at x=60°

Step-by-step explanation:

Given : Expression $3\tan x + \cot x= 5\csc x$

To find : The general solution of the given expression?

Solution :

$3\tan x + \cot x= 5\csc x$

$3\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}= 5\frac{1}{\sin x}$

$3\frac{\sin x\times \sin x+\cos x\times \cos x}{\cos x\sin x}= 5\frac{1}{\sin x}$

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

$3\sin^2 x+\cos^2 x= 5\cos x$

We know, $\sin x=1-\cos^2 x$

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

$3-3\cos^2 x+\cos^2 x= 5\cos x$

$3-2\cos^2 x= 5\cos x$

$2\cos^2 x+5\cos x-3=0$

$2\cos^2 x+6\cos x-\cos x-3=0$

$2\cos x(\cos x+3)-1(\cos x+3)=0$

$(2\cos x-1)(\cos x+3)=0$

$\cos x=\frac{1}{2},\cos x=-3$

We know, cos x lies between -1 to 1

So, We accept only $\cos x=\frac{1}{2}$

$\cos x=\cos 60$

So, x=60.

Therefore, The general solution of the expression is at x=60°