The given equation 2 c o s a . c o t 2 a − 6 c o s a − c o t 2 a = − 3 2cosa.cot^2a-6cosa-cot^2a=-3 2 cos a . co t 2 a − 6 cos a − co t 2 a = − 3 is identity equation.
Solving the above equation,
2 c o s a . c o t 2 a − 6 c o s a − c o t 2 a = − 3 2 c o t 2 a . c o s a − c o t 2 a + 3 − 6 c o s a = 0 c o t 2 a ( 2 c o s a − 1 ) − 3 ( 2 c o s a − 1 ) = 0 ( c o t 2 a − 3 ) ( 2 c o s a − 1 ) = 0 ( c o t a + 3 ) ( c o t a − 3 ) ( 2 c o s a − 1 ) = 0 2cosa.cot^2a-6cosa-cot^2a=-3\newline
2cot^2a.cosa-cot^2a+3-6cosa=0\newline
cot^2a(2cosa-1)-3(2cosa-1)=0\newline
(cot^2a-3)(2cosa-1)=0\newline
(cota+ \sqrt3)(cota-\sqrt3)(2cosa-1)=0 2 cos a . co t 2 a − 6 cos a − co t 2 a = − 3 2 co t 2 a . cos a − co t 2 a + 3 − 6 cos a = 0 co t 2 a ( 2 cos a − 1 ) − 3 ( 2 cos a − 1 ) = 0 ( co t 2 a − 3 ) ( 2 cos a − 1 ) = 0 ( co t a + 3 ) ( co t a − 3 ) ( 2 cos a − 1 ) = 0
Then, we get
c o t a + 3 = 0..............................1 c o t a − 3 = 0..............................2 2 c o s a − 1 = 0................................3 cota+ \sqrt3=0..............................1\newline
cota-\sqrt3=0..............................2\newline
2cosa-1=0................................3 co t a + 3 = 0..............................1 co t a − 3 = 0..............................2 2 cos a − 1 = 0................................3
General solution of 1 is a = n π + 5 π 6 a=n\pi +\frac{5 \pi }{6} a = nπ + 6 5 π , 2 is a = n π + π 6 a=n\pi +\frac{\pi }{6} a = nπ + 6 π and 3 is a = 2 n π + π 3 and a = 2 n π + 5 π 3 a=2n\pi +\frac{\pi }{3} \text{and}\space a=2n\pi +\frac{5\pi }{3} a = 2 nπ + 3 π and a = 2 nπ + 3 5 π .
Thus, the required solution are a = n π + 5 π 6 a=n\pi +\frac{5 \pi }{6} a = nπ + 6 5 π , a = n π + π 6 a=n\pi +\frac{\pi }{6} a = nπ + 6 π , a = 2 n π + π 3 and a = 2 n π + 5 π 3 a=2n\pi +\frac{\pi }{3} \text{and}\space a=2n\pi +\frac{5\pi }{3} a = 2 nπ + 3 π and a = 2 nπ + 3 5 π .
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