Question

If sum of all x in [0,2π] such that 3cot
2
x+8cotx+3=0 is Kπ then find K

Answer 1

Answer:

3cot

2

x+8cotx+3=0

Let cotx=y

3y

2

+8y+3=0

Let the roots be y

2

 and y

2

Sum of roots : y

1

+y

2

=

3

−8

Product of roots: $4y_{1}y_{2}=\cfrac{3}{3}=1$$

y

1

=

y

2

1

Since, y=cotx

∴cotx

1

=

cotx

2

1

​

cotx

1

​

=tanx

2

​

tanx

2

​

=tan(

2

3z

​

−x

1

​

)

So,

x

2

​

=

2

3z

​

−x

1

​

x

2

​

+x

1

​

=

2

3z

​

Value of K=

2

3

​ pls friend me

Answer 2

Answer:

k = $\frac{3}{2}$  

Step-by-step explanation:

3cot²x + 8cot x + 3 = 0

Let cot x = t

3t² + 8t + 3 = 0

$t_{1}$$t_{2}$ = $\frac{c}{a}$ = $\frac{3}{3}$ = 1 ⇒ $t_{1}$ and $t_{2}$ are reciprocal ⇒ $t_{1}$ = $\frac{1}{t_{2} }$ .... (1)  

$t_{1}$ = cot $x_{1}$ and $t_{2}$ = cot $x_{2}$ ----> (1)

cot $x_{1}$ = $\frac{1}{cot x_{2} }$ ⇔ cot $x_{1}$ = tan $x_{2}$

tan $x_{2}$ = tan ($\frac{3\pi }{2}$ - $x_{1}$) ⇒ $\frac{3\pi }{2}$ - $x_{1}$ = $x_{2}$ ⇒ $x_{1}$ + $x_{2}$ = $\frac{3\pi }{2}$

$x_{1}$ + $x_{2}$ = $\frac{3}{2} \pi$ ⇒ k = $\frac{3}{2}$