1.

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Answer»

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


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}}


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