Question

solve this ..

give proper answer☺️

Answer 1

Final Answer: x= $\frac{\pi}{20},\frac{\pi}{4}$

Steps:

$1) Let \:\: A=2x-\frac{\pi}{18},B=3x+\frac{\pi}{18} \\ => A+B= 5x$

2) We have,

$[cot(A)-1][cot(B)-1]=2 \\ \\ => [\frac{cos(A)}{sin(A)}-1] [\frac{cos(B)}{sin(B)}-1]=2 \\ \\ => [cos(A)-sin(A)][cos(B)-sin(B)]= 2 sin(A)sin(B) \\ \\ => cos(A)cos(B) +sin(A)sin(B) -cos(A)sin(B)-sin(A)cos(B)=cos(A-B)-cos(A+B) \\ \\ => cos(A-B)-sin(A+B)= cos(A-B)-cos(A+B) \\ \\ => sin(A+B)=cos(A+B) \\ \\ =>sin(A+B)=sin(\frac{\pi}{2}-(A+B))$

3) Since,Trigonometric Functions are Periodic.

So,

Case: 1

$sin(A+B) = sin(\frac{\pi}{2} -(A+B))\\ \\ => A+B= \frac{\pi}{2}-(A+B) \\ \\ =>2( A+B) = \frac{\pi}{2} \\ \\ => A+B = \frac{\pi}{4} \\ \\ => 5x= \frac{\pi}{4} \\ \\ =>x= \frac{\pi}{20}$

4) Case :2

$sin(A+B)=sin(2\pi+\frac{\pi}{2} -(A+B)) \\ \\ => 2(A+B)= 2\pi +\frac{\pi}{2} \\ \\ => 2*5x= 2\pi +\frac{\pi}{2} \\ \\ => x= \frac{\pi}{4}$

Other cases will not not give solutions in restricted x such that

$0\leq x \leq \frac{\pi}{4}$

Hence , Final result is

$\boxed {x=\frac{\pi}{20},\frac{\pi}{4}}$