The general solution of sin^2 theta sec theta | vedclass

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Question

(B) The given equation is $\sin^2 	heta \sec 	heta + \sqrt{3} 	an 	heta = 0$.Since $\sec 	heta = \frac{1}{\cos 	heta}$,we can write the equation

Vedclass · Accepted Answer

$	heta = n\pi, n \in Z$

Vedclass · Answer

(B) The given equation is $\sin^2 	heta \sec 	heta + \sqrt{3} 	an 	heta = 0$.Since $\sec 	heta = \frac{1}{\cos 	heta}$,we can write the equation as:$\frac{\sin^2 	heta}{\cos 	heta} + \sqrt{3} 	an 	heta = 0$Using $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$,we have:$\sin 	heta 	an 	heta + \sqrt{3} 	an 	heta = 0$Factoring out $	an 	heta$:$	an 	heta (\sin 	heta + \sqrt{3}) = 0$This gives two cases:$1) 	an 	heta = 0 \Rightarrow 	heta = n\pi, n \in Z$$2) \sin 	heta = -\sqrt{3}$. Since the range of $\sin 	heta$ is $[-1, 1]$,$\sin 	heta = -\sqrt{3}$ has no real solution.Therefore,the general solution is $	heta = n\pi, n \in Z$.

The general solution of $\sin^2 \theta \sec \theta + \sqrt{3} \tan \theta = 0$ is

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