If tan^2 theta - (1 + sqrt{3}) tan theta + sq | vedclass

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Question

(A) Given the equation: $	an^2 	heta - (1 + \sqrt{3}) 	an 	heta + \sqrt{3} = 0$Factor the quadratic equation:$	an^2 	heta - 	an 	heta - \sqrt{

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$n\pi + \frac{\pi}{4}, n\pi + \frac{\pi}{3}$

Vedclass · Answer

(A) Given the equation: $	an^2 	heta - (1 + \sqrt{3}) 	an 	heta + \sqrt{3} = 0$Factor the quadratic equation:$	an^2 	heta - 	an 	heta - \sqrt{3} 	an 	heta + \sqrt{3} = 0$$	an 	heta (	an 	heta - 1) - \sqrt{3} (	an 	heta - 1) = 0$$(	an 	heta - \sqrt{3})(	an 	heta - 1) = 0$This gives two cases:$1) 	an 	heta = \sqrt{3}$ $\Rightarrow 	an 	heta = 	an(\frac{\pi}{3})$ $\Rightarrow 	heta = n\pi + \frac{\pi}{3}$$2) 	an 	heta = 1$ $\Rightarrow 	an 	heta = 	an(\frac{\pi}{4})$ $\Rightarrow 	heta = n\pi + \frac{\pi}{4}$Thus,the general solution is $	heta = n\pi + \frac{\pi}{3}, n\pi + \frac{\pi}{4}$.

If $\tan^2 \theta - (1 + \sqrt{3}) \tan \theta + \sqrt{3} = 0$,then the general value of $\theta$ is

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