The general solution of the equation 2 tan th | vedclass

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(B) Given equation: $2 	an 	heta - \cot 	heta = -1$Since $\cot 	heta = \frac{1}{	an 	heta}$,we have:$2 	an 	heta - \frac{1}{	an 	heta} = -1$

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$n\pi - \pi/4$

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(B) Given equation: $2 	an 	heta - \cot 	heta = -1$Since $\cot 	heta = \frac{1}{	an 	heta}$,we have:$2 	an 	heta - \frac{1}{	an 	heta} = -1$Multiply by $	an 	heta$ (where $	an 	heta 
eq 0$):$2 	an^2 	heta - 1 = -	an 	heta$$2 	an^2 	heta + 	an 	heta - 1 = 0$Factor the quadratic equation:$(2 	an 	heta - 1)(	an 	heta + 1) = 0$This gives two cases:Case $1$: $2 	an 	heta - 1 = 0 \implies 	an 	heta = 1/2$Case $2$: $	an 	heta + 1 = 0 \implies 	an 	heta = -1$For $	an 	heta = -1$,the general solution is $	heta = n\pi - \pi/4$,where $n \in \mathbb{Z}$.

The general solution of the equation $2 \tan \theta - \cot \theta = -1$ is

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