find the general solution of tanx + cotx = 2cosecx

We have,

tan(x) + cot(x) = 2 cosec(x)

⇒ [ sin(x) / cos(x) ] + [ cos(x) / sin(x) ] = 2 cosec(x)

⇒ { sin2(x) + cos2(x) } / sin(x) cos(x) = 2 cosec(x)

⇒ 1/sin(x) cos(x) = 2 cosec(x)

⇒ 1 / sin(x) cos(x) = 2 cosec(x) = 2 / sin(x)

⇒ sin(x) cos(x) = sin(x) / 2

⇒ 2 sin(x) cos(x) = sin(x)

⇒ cos(x) = 1/2 = cos(/3)

⇒ cos(x) = cos(/3)

⇒ x = 2n ± (/3), where n ∈ Z

Hence, the general solution is given by x = 2n ± (/3), where n ∈ Z.

We have,

tan(x) + cot(x) = 2 cosec(x)

⇒ [ sin(x) / cos(x) ] + [ cos(x) / sin(x) ] = 2 cosec(x)

⇒ { sin^2(x) + cos^2(x) } / sin(x) cos(x) = 2 cosec(x)

⇒ 1/sin(x) cos(x) = 2 cosec(x)

⇒ 1 / sin(x) cos(x) = 2 / sin(x)

⇒ sin(x) cos(x) = sin(x) / 2

⇒ cos(x) = 1/2

⇒ cos(x) = cos(/3)

⇒x = 2n +/- (/3),where n ∈ Z

Hence, the general solution is given by x = 2n +/- (/3), where n isin Z.

What's happening? I typed pi symbol in the above answers but it is not showing. Neither I'm able to make texts bold or italic nor able to show "square" (^2) in the above answers.

tan(x) + cot(x) = 2 cosec(x)

⇒ [ sin(x) / cos(x) ] + [ cos(x) / sin(x) ] = 2 cosec(x)

⇒ { sin^2(x) + cos^2(x) } / sin(x) cos(x) = 2 cosec(x)

⇒ 1/sin(x) cos(x) = 2 cosec(x)

⇒ 1 / sin(x) cos(x) = 2 / sin(x)

⇒ sin(x) cos(x) = sin(x) / 2

⇒ cos(x) = 1/2

⇒ cos(x) = cos(/3)

⇒x = 2n +/- (/3),where n ∈ Z

Hence, the general solution is given by x = 2n +/- (/3), where n ∈ Z.