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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

The solution of $\tan 2\theta \tan \theta = 1$ is

The general solution of $\sin x + \cos x = \min_{a \in \mathbb{R}} \{1, a^2 - 4a + 6\}$ is

Find the general solution of $\csc x = -2$.

The number of solutions $x$ of the equation $\sin(x+x^2) - \sin(x^2) = \sin x$ in the interval $[2, 3]$ is

The smallest positive angle which satisfies the equation $2\sin^2 \theta + \sqrt{3} \cos \theta + 1 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module