Questions & Answers

Question

Answers

A) $n\pi + \dfrac{{4\pi }}{3}:n \in I$

B) $n\pi + \dfrac{{2\pi }}{3}:n \in I$

C) $2n\pi + \dfrac{{4\pi }}{3}:n \in I$

D) $2n\pi + \dfrac{{2\pi }}{3}:n \in I$

Answer
Verified

In this we have to remember the basic coordinate system of all the trigonometric function and their basic general solution formula

Here we choose first equation $\tan \theta = \sqrt 3 $

And solution for $\theta = \dfrac{\pi }{3},\dfrac{{4\pi }}{3},\dfrac{{7\pi }}{3}.....$

Now we take second equation $\cos ec\theta = - \dfrac{2}{{\sqrt 3 }}$

So we convert this in term of $\sin \theta $

By using formula $\cos ec\theta = \dfrac{1}{{\sin \theta }}$

So we have $\dfrac{1}{{\sin \theta }} = - \dfrac{2}{{\sqrt 3 }}$

And $\sin \theta = - \dfrac{{\sqrt 3 }}{2}$

Now we know that $\sin \theta $ is negative in $IIIrd$ and $ivth$ coordinate

So $\theta = \pi + \dfrac{\pi }{3},2\pi - \dfrac{\pi }{3}$ this value is repeated after every $n\pi $ interval.

$\theta = \dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3}$

Now the common value in both the equation is $\dfrac{{4\pi }}{3}$

So the general solution is $n\pi + \dfrac{{4\pi }}{3}:n \in I$

For graphical method have follow this steps

* Make graph of each function

* Denote all the values and compare where both take the same value.